5- Compilado Ejercicios de Ec. de Energías

efectos nada de la unidad ventilador-motor que se usará puede tomarse5.109 27 m como 50 por ciento. Si el ventilador20 debe reemplazar todo 5.105 5–81 Water A water flowssiphon having at a rate of 20 a constant inside L/s through diameter of a horizontal sure at m el volumen de aire en 10 min, determine a) la potencia de la 3 in. is pipe arranged whose as shown diameter is in Fig. constant P5.105. at 3 If cm. the The friction pressureloss dropbe- minimi o se re- tween A and across a valve the2pipe B is in0.8V /2, where V is the is measured to velocity be 2 kPa.ofDetermine flow in the piping, unidad motor-ventilador que debe comprarse, b) el diámetro siphon, determine the the irreversible headflowrate loss ofinvolved. the valve, and the useful pump- losses i   del ventilador y c) la diferencia deCompilado   presión de uno a otro lado ddee  Energía:  Cenguel,   ejercicios   ing powerWneeded hite,  Mto unson.     resulting pressure drop. overcome   the minimi 3 este último. Tome la densidad del aire como 1.25 kg/m yIng.  des-Waldo  Lizcano.   elaciona   carte el efecto de los factores de corrección FIGURE P5–78 de la energía ciné- Answers: 0.204 m, 40 W   5.110 tica. Section inética? Nota:   4 ft shock a 5–79 A hydraulic turbine has 85 m of head available at a shock a 𝑊!"#$% = 𝑚 𝑔flow ∗ ℎ!"#$% = 𝜌m∗3/s,𝑉 and rate of 0.25 ∗ its 𝑔 ∗overall ℎ!"#$%   𝐾𝑤   turbine–generator effi- A Determ riba del 8 m/s the cha 𝑊!"#$%&' = 𝑚 𝑔 ciency ∗ ℎ!"#$%&' = 𝜌 ∗ Determine is 78 percent. 𝑉 ∗ 𝑔 ∗ the ℎ !"#$%&' 𝐾𝑤   output of electric   power 12 ft a boqui- this turbine. Ventilador 4 ft ton and directo   de extracción 5–80 The demand for electric power is usually much higher Aire e por la during the day than it is at night, and utility companies often é puede sell power at night at much2/17/12 lower prices to encourage con- del nivel c05FiniteControlVolumeAnalysis.qxd BMAns.qxd 2/15/12 4:00 9:01 PM PM Page Page 270ANS-3 A)  Flujos  con  Pérdidas  y  Ganancias  de   sumers to use the available power generation capacity and to Water 20 L/s avoid building new expensive power plants that will be used 3 in. 1 na bom- Energía   only a short FIGURA P5-73time during peak periods. Utilities are also will- ing to purchase power produced during the day from private B ∆P = 2 kPa ia hasta parties at a high   price. c05FiniteControlVolumeAnalysis.qxd 2/17/12 4:00 PM Page 270 e dicha 𝑃! 𝑉! ! Suppose a utility company is selling electric power for ■ Figure P5.105FIGURE P5–81   dmisión 5-74 + Se 𝛼 está + bombeando 𝑍!$0.03/kWh +ℎ agua desdeand un islago grandeto hasta un P máximo 𝜌𝑔 depósito 2𝑔 está 25 m!"#$% que at night arriba, a razón de willing 25 L/s, a pay $0.08/kWh través de for   Cha power produced !during the day. To take advantage of this 5.106 Water flows through do de la una bomba 270(potencia Chapter opportunity, 5 𝑉■! Finite en𝑃!la flecha) de 10Control Volume kW.isSiconsidering la pérdida Analysis irre- 5–82E The water level ainvalve Answersa tank (see Fig.ftP5.106) toisSelected 66 above at the Even-Numbered the rate of Homewor ground. = + 𝛼an entrepreneur + 𝑍! + ℎ !"#$%&' building + ℎ!é!"#"$%   a large 1000 lbm/s. The pressure just upstream of the valve is 90 psi and 1-in. dia A hose is connected to the bottom of the tank at the ground entrada versible de carga reservoir del sistema 𝜌𝑔 40 m de2𝑔 tuberías above the es de level, lake 7 m, pumping determinewaterla from the the2level pressure drop acrossatthethevalve is the 50 psi. s de co- eficiencia0.75 mecánica m de la bomba.   Respuesta: 78.5 4.32 720 m!s2; 1440 m!s por ciento ; 2160 the 2nozzle andm!s end of 5.76The hose inside diameters 1b2 is pointed 234straight lb of the up. valvetank The inlet andisexit cover pipesbutarethe12pressure airtight, and 24over in. If the flow 2 4.34 !33.8 ft!s ; 1.05 through the valve occurs in a horizontal plane, 9170the 5.78 determine lb water 5-75 270de EES 𝑚 Vuelva a1.0considerar = 𝜌𝑉   el problema mChapter 5 ■ Finite Control 5-74. Use 4.36Volume el Soft- 1225x 2 surface " 1502 m!sin; available Analysis 0; 150 m!s is unknown. 2 ; 375 Determine 2 m!sthe valve. the minimum 5.80from78.5 tank air kN to the loss pres- ware (o cualquier otro programa de este ti- energy across   Reservoir 4.42 3m 8.02so-ft!s sure (gage) that will cause a water stream the nozzle 5.88 1a2 231 N & m, 185   rad!s po) e investigue el efecto de la pérdida irreversible de carga rise 90 ft from the ground.12 in. 2 2 1.5 m 𝑊 = 𝑚𝑔ℎ   6m 4.46 1c2 116 N & m, 92.5 rad! bre la eficiencia 0.75mecánica m de la bomba. Suponga 223 que la 8860 pérdida m!s ; 7440 m!s 5–83 A large tank is initially filled with water 2 m above de carga varía de 0 hasta 15 m,  en incrementos CAPÍTULO de 54.48 1 m. Trace 1a2 10laft!s2; 1b2 20 ft!s 2 , 10 the center ft!s 2 ; 1c2 22.4 of a sharp-edged ft!s 2 5.90 10-cm-diameter orifice. The 39.6 hp tank 2 2 2 2 gráfica de los resultados 1.0 y analícelos. m 4.50 40 m ax # 2c x 1x " y 2; awater y # 2c y 1x2 " y22 to the atmosphere, 5.92 1a2 43 deg; 1b2 53.4 ft & l superfi- agua de éste hacia ese depósito   en la noche, usando 4.52 3 energía 3.20 m!s2; 1.48 m!s2drains24toin.the atmosphere. If the12total surface is open and the orifice onduce ■ Figure P5.103 m 5.92 !36.8 in. irreversible head loss ft & lb!slug in  barata, 5-76 ySedejar usa que una el bomba agua de 7 hp fluya del (potencia depósito 6m en regreso de la flecha) 4.54 al lago para 2 the2 system is 0.3 2m, determine the initial !225 ANS-3ft!s ; !28.1 ft!s ; !8.33 ft!s 5.96discharge ft & lb!slug 1.5 m !3130velocity la pre- subir agua durante el día, hasta produciendo BMAns.qxd una altura 2 potencia de 2/15/12 15 m.cuando Si9:01laPump– eficiencia laPMunidad Page mecánica bomba- tura de A.1)  Pérdidas   de la bomba 5.104 motor opere como un turbogenerador en el desarrollo del flujo es de A siphon 82 por is ciento, used to determine draw water el turbine at gasto 4.56 70 !F 525 from volumétrico a $F!slarge con- of water from the factor at the orifice to be 1.2. tank. Take the 5.100 kinetic energy ■ Figure P5.106 3 0.32 correction ft; right to left de pre- máximo Un inverso. de tainer agua. as indicated in Fig. P5.104. The inside análisis preliminar   muestra que puede 4.58 diameter usarse !28.4 of the siphon un $C!s; !25.1 $C!s 5.102 0.042 m !s line ■ is 1Figure in. and P5.103the pipe centerline rises 34.60 ft above 1a2the 200 essentially $C!s; 100 5.107 $C!s5–84 Water in.enters a hydraulic turbine 5.104 through 4.58 aof %30-cm- 10!3 m3!s m3/sun entubo cualquiera A gas 24 gasto 5-77 deFluye agua de 2water tank.de las dos direcciones, y3 length agua en horizontal, cuyo Lake expands through a nozzle from a pressure 300 psia constant level in the Show thatdiámetroby 4.62 varying 5.0 sethe re- of !s for eachtoline diameter pipe at a rate of 0.6 m 3/s and exits through a 25-cm- laduce pérdida de 15 the irreversiblecm siphon hasta below 8decm carga themediante water del level, sistema de rate un h,reductor, the tuberíasaofrazónflow es m de de3 through the a pressure of 5 psia. The enthalpy change diameter pipe. The pressure drop in the turbine is measured 5.106 involved,5660 ȟft1 " & lb!slug ȟ2, is 40.035 m. Se 5.104 A siphon is used to draw water 4.64 at 70 132 from ft !s a large con-150 Btu/lbm. If the expansion is adiabatic 5.108 but with 926 ft frictional & lb!slug; effects 200 ft & m3espera siphon /s. Si seque can mide be laschanged. que eficiencias la presión Assuming combinadas la líneadecentral en frictionless laflow,bomba- !F es determine de the by a mercury manometer to be 1.2 m. For aFigure ■determine combined P5.106 turbine– exit  gas motor y del tainer as turbogenerador indicated inde Fig. 75P5.104. theThe 4.68 inside 1b2limiting diameter 2Se V0ofhb!3 the siphonand generator the inlet gas speed is negligibly small, 5.110 734net the ft electric 470 kPa maximum y 440 kPa, flowrate antes FIGURE yseadespués possible P5–80 through por del ciento reductor, siphon. cada The una. respectiva- condi- velocity. efficiency of 83 percent, determine the supone mente, determine que tion el line is sistema the is la pérdida 1 occurrence in. opera and the durante irreversible of pipe cavitation centerline 10 h de Showin en rises the carga that 4.72 los 3 siphon. ft modos en éste. above 1a2 Will 3000 de the the essentially kg!s; actual Tome the length of 5.107 Answers 1b2 3.00 m 3 !s A gas expands   throughEven-Numbered to Selected 5.112 a nozzle from a pressure 2.22 % Homework 10 6 of 300 W psia Problems bomba y demaximum constant turbina, flow cada water be uno,morelevel en orin the less un tank. than día the normal; frictionless by determine varying value? el Explain. 5.108 For the 180! elbow and 5.114 nozzle flow 5.48shown hp in Fig. los factores de to a pressure of 5 psia. The enthalpy change involved, ȟ ȟ2, is the corrección de the la water energía cinética h, the como rate of1.05. 1 " siphon below level, flow through theP5.108, determine the loss in available energy from section (1) to ingreso Respuesta: potencial 0.68 m que este sistema de bomba-turbina Chapter 5 puede 2   150 Btu/lbm. If the expansion is 5.116 adiabatic but Will with work frictional for 2 atm., no effects e el gas- siphon can be changed. 4.32 Assuming 720 m!s frictionless 2 ; 1440 flow, m!s determine ; 2160 m!s thesection 2 5.76 1b2 234small,lb determine generar por año. 5.2 20 m!s and(2). theHow inletmuch additional gas speed available is negligibly energy 5.118 is lost m 0.052 from3thesec- !s exit gas do de la 5-78 El nivel maximum del agua flowrate en unpossible   tanque 4.34 through está !33.8 the m 20 siphon. ft!s 2arriba ;siphon. 1.05 Thedel limiting condi-tion (2) to where the water comes to rest? velocity. 5.78 9170 lb5.120 1610 ft & lb!slug; 1110 f 5-81 suelo. Fluye Se conecta tion ispor agua the un una occurrence manguera tubo horizontal, of 3al cavitationdel ft fondocuyo in5.4 the diámetro tanque 3.66 slugs!s y Will 2es la the actual2 de tube- 4.36 1225x " 1502 m!s ; 0; 150 m!s ; 375 m!s 2 5.80 78.5 kNnozzle1a2flow constante de 3maximum cm, en a razón flowde be 20 more L/s. or Se less than the 5.6 quefrictionless 1.57 in. value? Explain. 5.108 6 in.For the 180! elbow and5.122 4.08 shownhp; 1b2 in 9.94 Fig. ft boquilla que está el extremo de dichamide manguera la se caídaapunta de 4.42 8.02 ft!s 5.8 4.99 ft!s P5.108, determine the5.88 loss in 1a2 2315.124 available Nenergy & m,0.58 185 fromrad!s; section1b2(1) 200 to N & m presión a través de una válvula en el tubo es de 2 kPa. 2 Determi- 2 4.46 8860 m!s 5.10 ; 7440314 m!s m!s section (2). How much additional 1c2 available 116 N 5.126 & energy m, 92.5 1.78 iskW lost from sec- rad!s ne la pérdida irreversible de carga de la válvula y la potencia 2 útil de bombeo necesaria para vencer 4.48 la 3caída 1a2 10 5.122; 1b2de ft!s resultante !0.0189 20 pre- ft!s ,kg!s 10 ft!s2; 1c2tion 22.4 (2)ft!s to where 2 the water 5.90 comes 39.6 to rest? y hp 5.128 1a2 0.56; 1b2 11.7 lb ft 2 2 2 32 2 Section 2 (2) sión. Respuestas: 0.204 m, 40 W h 4.50 a # 2c5.14 x 1x 0.0125 " y 2; a lb # !ft2c y 1x " y 2 5.92 1a2 43 deg; 5.132 1b22.3653.4ftft & lb!s m23  !s x ym 6 in. 4.52 3.20 m!s 5.16 2 ; 1.48 1.00 m!s 12 in. 5.92 !36.8x5.134 ft & lb!slug 1a2 17.2 deg, 4.29 m!s; 5 4.54 !225 ft!s 5.182; !28.1 150 liters!s ft!s2; !8.33p ft!s 2 5.96 !31305.136 ft & lb!slug 1a2 3.43 ft & lb!lbm; 1b2 3 1 = 15 psi 1 in. 4.56 525 $F!s 5.20 3.63 ft!s V1 = 5 ft/s 5.100 0.32 5.138 ft; y right Nota:  T 11.6 to ubería   left Nel  &plano   en   m!kg; 0.796 4.58 h !28.4 5.22 $C!s; !25.1 0.711; $C!s 0.791; 0.837; 0.866 Section (2) 5.102 0.042 horizontal   5.140 m 3 !s y  descarga   148 ft & el   lb!slug; 0.875 ponible Section (1) ambiente   x bogene- 4.60 1a2 2005.24 $C!s; 100 1a2 6280 $C!skg!s 12 in. 5.104 4.58 % 10 m !s !3 3 ■ Figure P5.108 ia eléc- ■ Figure P5.104 4.62 5.0 m35.26 !s for each 1a2 0.00456 line slug!s increasing; 5.106 5660 Chapter ft & lb!slug 6   1 in. 4.64 132 ft !s 3 1b2 2.28 % 10 !4 slug!ft 5.109 p 13= & An 15 s V1 = 5 ft/s psi automobile   5.108 926 ft 6.2 engine will work best when & lb!slug; the 2x, 200 back2 4xt ;ft!2y, pres- & lb!slug 4yt2; V # 5.105                                3      in.        is        arranged A water siphon                4.68                          as        shown having    in     Fig.1b2 a constant 2 V0 hb!3 5.28 1260 inside min diameter of sure at the interface of the exhaust 5.110   Section manifold 734(1)ft and the engine a # 5.66 ft!s block is 2 ho más P5.105.5.30 If the 14.8 m!s loss 3be- minimized. Show how reduction of losses 6.4 in the 6 exhaust manifold, s com- 1  En  algunos  ejercicios  se  tiene  en  cuenta   2 4.72 e l   f 1a2 actor   d 3000 e   c kg!s;dfriction orrección   1b2 e   l a   3.00 m !s piping, ■ Figure and P5.108 muffler will 5.112 also reduce   2.22 the % back 10 W pressure. 1a2 0; 1b2 How v # !1y ' 2 " z could tween ■A Figure and B P5.104 is 0.8V /2, where V is the5.32 velocity 1a2 of 15.6 flow in gal!min;the 1b2 62.4 gal!min a en la Energía  cinética:   siphon, α  yAgua determine  el  Término  the para   flowrate EK  se  convierte   involved. en  𝑬5.34 𝑲 = 𝜶 𝟐120 𝑽𝟐 .         N losses in the exhaust system 5.114   5.48 hp be reduced? No What primarily limits 2 the nsumi- (α  =  2  F.  Laminar)  20     L/s Chapter 5 5.109 An automobile5.116 3 minimization of exhaust system engine   losses? will work Will 6.6 best work for zwhen 2 #atm.,3xy thenotk̂back ; No forpres- 3 atm. 5.105 A water siphon having a5.36 constant 2.66 inside % 10 !4 diameter m !s of sure at the interface of the exhaust manifold and the engine 3 block x 3 is (1,04  <  α  <  1,11  F.  Turbulento    ó  α  ≈1)  5.2 20 m!s 3 de po- 5.118 0.052   article m !s loss be-5.110minimized. in“Smart !Shocks,” (SeeShow Fluidshow in the News titled 2 3 pin. 5.38 If15.5 lb reduction 6.8 of losses u #exhaust the x manifold, " " f 1y2 radoras  Ver  Pag.  208  Cengel   ara  is arrangeden  asel  shown profundizar   =t2ema.   5.4 in Fig. P5.105. 3.66Vslugs!s the friction 5.120 1610 ft lb!slug; 1110 2 How 3 ∆P0.8V kPa 2 3 !s; theSection 5.3.3.) andAmuffler 200-lb force applied to the &end of pressure. the piston offtthe lb!s & could los pe-   tween A and B is /2, where is5.40the velocity 7.01 ftof flow674 in lb piping, will also   causes the two reduce the back 2  Se  recomiendan:   siphon,5determine -­‐81;    5.108;   the flowrate 5.6  P3.133   4 ft 1.57 involved. in.5.42 352 lb shock absorber losses in theshown exhaust in system Fig. 5.122 P5.110 be1a2reduced? 4.086.10hp;What 1b2g# primarily 9.94 # ft !r0of ends vlimits the the omprar FIGURA P5-81 5.8 4.99 ft!s shock absorber minimization to move of toward exhaust 5.124systemeach other losses? 0.58 with a speed r of! 5 r ft/s. as a un A 5.44 2.69 lb Determine the head loss associated with the flow of the oil through 0 i 5.10 314 m!s 5.46 1a2 6.26 m!s; 1b2 392 5.110 N; the channel. 1c2 675 NeglectN(Seegravity Fluids 5.126 in the and any 1.78 News kW6.12 article friction force No titled “Smart between theShocks,” pis- 12 ft 5.12 !0.0189 5.48 kg!s 0 N ton Section and cylinder5.3.3.) A walls. 200-lb 5.128 force 1a2 applied 6.14 0.56;to 1b2 the end1a2 11.7 2.24 oflbthe m!s, piston of the deg !63.4 público 4 ft 5.14 40.0125 ft lb 5.50 !ft 3 482 N downward shock absorber shown in 5.132 Fig. P5.110 2.36 ft causes the 76.0 two deg, ends none of the y desea 5-82I El nivel del agua en un tanque está1.0066 ftmarriba m del sue- shock absorber to move toward each6.16 other with a speed of 5 ft/s. el día. A 5.16 3 5.52!s 17,900 lb 5.134 1a2 17.2 deg, No 4.29 m!s; 558 N & m!s lo. Se conecta una manguera al fondo del tanque a nivel del pi-3 Determine the head loss associated with the flow of the oil through 1b2 392 N. Air 5.000 gal/min in either direction.2 20 m!s P3.3. flow rate Q m /h will result? 5. p1flow.866 of exhaust system losses? (a)ftIn a5.5. P5. the length News L. P5. P5.24 m!s.34 !33. 2x top to the # a hydraulic c ! 4y " C. m3/sEven-Numbered Hom adds to Section 5.6y ! 100 kPa.1 ft!s . ln r " Br co percent.122 State fluid use the that from 4. 3-ft  5.42 8. 4yt2. The water3 density ft tempera- a # 5. For a design flow rate 5.64 132 ft3!s 6 in.117.54 psi and28.36 f # and Fig. where V is 5.60 1a2 200 section $C!s.110 72 percent efficient and is driven by a 500-W motor.107 2 A gas expands 2 through2a 2nozzle from a 2pressure of 300 psia3 constant water level in the tank.11512 in. Turbine ■ Figure P5.8 voirs to restore the situation.43 ft & lb!lbm. h.115 5. in 5.135 draws water from P siphon.900 of lbChannel h 6. !8. Figure P5.05 †5. 20–2000Q2 hp =P3.10 # 14!32 6.determine P5.38 60.132 shown inlet pipe   in Video V5.134 A 36-in-diameter pipeline carries oil (SG 0. !28.26loss is 1a2 0. 20 4.34pipe120 at a N rate of 6. If the pump is 4-in.133rises 3 ft with is filled abovewater the essentially at 20°C.12 No 5. If the differential Water at 20˚C 5. human 2 5. The inside 5. Repeat gacross #the prob.72 1a2 3000 kg!s.74 1a2 181 The the hydrant.96 lb 6. 2 5 20 5. P3.   The inside diameter of the siphon 5.122                                Water                Water  is      supplied          is      pumped        at        150        from    ft    3/saand tank. If the friction loss be.114 lb the Will flowrate a pump Theof pumper that water adds istruck 3 20hpftto 3 shown/s.8 lb.4 siphon iscanopen and waterAssumingflows at 500 m /h.6 1.26 m!s.132 is used to draw   water at 70 !F from a large con- 4.117  5.16 as indicated 6.54 !225 ft!s velocity. shown in5. the flow remains 6 in.02 lb 20 ft L !y 2 evated of the valvetank. turbine V2 !is010-in.133 in Fig.138Students 1 = 5 ft/s mercury amount less? of manometer loss mechanics is associated is 3 laboratory ft. the upper reservoir in the daytime to produce power for a 5.5.5 ft3/syour (a)to the aanswer maximum # !rthe0v Upstrea 5.100 How $C!s much additional available energy is lost 5. P5.136 60 ft the A pump is pressure dropto deliver water at 20°C H from a pond to an el- Oilacross the valve is 50 psi.711.119 5. If the pump efficiency5.94 shaft the flowing droelectriccondition? turbine shown in Fig.121 6.120 5. of the valve is 90 5.122 atuniform The a rate P5. 0. if the sections (1) average and (2).136 could P3.69 lb lem if theand tank the flowrate pressurized expected to 3.24of 1a2 6280 5.111 Based on flowrate and 5. calculate the power5.62 5.14 Esti.2 Application of the Energy Equation— pressure 6atftthe 4-in. It is planned to ■ pumping power required. surized tank asHshown in Fig. m 0.96 !3130 ft & lb!slug maximum flow be more or less than the frictionless value? Explain.36 reading 2.126 Fig.57 in theoffluid output Why the actual amount of 9. 3 1. noneSection 5.134 1a2 17.(2).110 kg!s experiment. 1b2 head 20 is ft!s h p ! . is the siphon below the water level.5.123 With Shaft Work losses are negligibly small.theaflowrate.6 ft /s.A. m3is/s to deliver adds to static pressure the water at section if the head (2). hp Deter- affect the When valve A is closed. T 6.3.34 point psi to(1).22 reser.16 m3!sat the interface of the exhaust manifold and the engine block is 5. 1. ȟ2.88 1a2 231 Nhy-& m.104   ft of pipe.The Hg va 2 Z 2 = 25 ft turbine. for 3.106 5660 ft & lb!slug 3 ft Pump 10 ft 4.90 Q is 203 in 39.138 #limits gd2). The p =675 0 N head loss is known to be 1.66 pslugs!s 85.01 ft3!s.   (2).105the pipe.0. as in Fig.33 5. in Fig.S. of losses entrance in the losses are negligi- exhaust manifold.60 !185 kN.6 z # 3xy2 k̂.  Munson.100 0.92 effects 1a2 43 deg. another m3as!s (2) passes ■ isFig lo What is the friction head loss between 1 and 2. 5.4the pis.4 $C!s.00 mP5.106 Water flowshthrough a valve (see Fig.110 734 ft ■ Figu 2 4. f # aerator power (a) (b) column. in the18 5. 1.Pipe area = 0.108 926 ft271 Problems & lb!slug.112?   Vin.7Pump lb 2x through the valve in a horizontal determine the loss psi 12 in.3. 0. what horsepower motor is required 0 to drive it? P3.possible? P5.791. 92.88) flows in an inclined 5.5pumplbsup. minimization V is the0.133 h A Chapter 5 5.36 1225x " 1502   m!s the power . 606. 1b2 v #■!1y ■Figu Fig'2 1 Z = 150 ft 5.22 12 in.63 ft!s losses in the aexhaust through long 600 ft pipe system is given be reduced? by hf. when the Because back pres- 5.12 !0. 2°C. estimate amount of lion barrels per day (bbl/day) (1 bbl ! 42 U.8 ■ Figure P5.loss = 15 psi amount 5 ftin/s available as power shownenergy in Fig.137 A fireboat draws seawater (SG ! 1. Theand inlet pump exitis pipes 1 m above are 12 theandpond. ■ Figure P5.68 66. The Water hydroelectric flows by gravity turbinefrom shown onein5. P3. (Seediameter Fluids Oil in d. bine and (b) Adelivered by the pump. (b) the of appropriate pressure 60 ft above calculations. 4xt2.114 work are for is to deliver negligible. # p2determine c05FiniteControlVolumeAnalysis.102 (1) to 0.106 large power tankoutput shown ofintheFig. What is the maximum possible power output of the1 hy.66 2.114 The pumper 0 5. where across c ! 0.0 m !s for each tion 3 (2) to where the water Q line V comes to rest? 5. 5. the 11.140 Section average speed through the pipe.108 of temperature.8 kN 200 6.ofinclined 80 What is pipe gpm. !25. 1a2 0. determine 1b2 Yes. 4.76 1b2 234 lb column 4.113. inAFig.1 $C!s 50themloss in available energy from section 5. .38 that the15.58 in the 1a2 W1 ! 9.long and pipe the pipe centerline P5.46 1a2 6.Aincreasing.32 1a2 15.66 in % the 10!4 m■3!s Figure P5.108. determine the exit gas maximum flowrate possible through the siphon. It will be shown in Chap.2 Application of the Energy Equation— (b)pressu Det (a) (b) With Shaft Work losses   4.138 P5.8 ft!s2. ■ Figure P5. Estimate the power 4inft kW (a) extracted by the tur.116 Will work  forenergy 2 atm †5. 2160 2 m!s 2 3 5. determine the flowrate involved.20 3.89) at 5.138.48 m!s 2 5.042 m3!s diameter 13 ft 120 ft/s 4.) and 200-lb volume forceflow applied ratetoQthe is end measured of the piston with Chapter of the6 the 17 ft.20 and m!sthe 2 .07 m Answers   to Selected Q.112 at a hig P3.7 city.121 The turbine shown in Fig.56 diameters 4.120 gal!min No 4 ft1 ■ Figure P5. article andH water titled level “Smart height Shocks.8 1 mil.113.66 ft!s2 Determine the head loss associated with the flow of the oil through Turbine 5. right to left Constant- 4. !63. 1b2 11.114 that5.  Waldo  Lizcano.48 the hp pump Section (2) must add to the water.3 lb 3  turbine Hydrant Sof e  raecomiendan:   water through plantaaerator.4 P5. ■ Figure The total head loss P5. The enthalpy change involved. 5.0 m Penstock 5. 1b2 sup- lower 5. P5.135 between the pump head as shown Fig.with calculate pumping at Penn the power the 5.48 1a2 10 ft!s . it pumps water from lower to upper 5.68 1b2 2 V0 hb!3 5. 1b2 53.3.112? 2 1c2 116 N & m. what 4.115 Fig.5 ft /s to a maximum elevation of 60 ft above 5.14 1a2 2.0 deg.117 24-in 3 mercury manometer is 3 ft. m3/s.62 what 1290 lb!ft.052 lake P5.113is 6. 3120 lb!ft diameter Free jet !Ar 2 3 6. this purpose? 5. c05FiniteControlVolumeAnalysis.111 Based on flowrate and pressure rise information.10 The 314 m!s function Figure P5. 0.05 0.”H 15.5 m Ing.0 discharge ftQ3/s.4 m. " yIf2the 2.08 pump the hp.5 kN plant is to 5. 45.6 N & m!kg. 2x. and muffler will h also reduce the back pressure.40 of1a2theYes.104 A siphon P3. OilP5. 3 2 Determ pres.     3m   0 of a wa 12 in.52 3 4 P3.116 at a rate of 1000 gal in 10 min 2de. the relationship Piston 76.114 Q.118 Aerator p1 = 60 psi column (2) .113 sketched million 3 GO ingal/min Fig. (c)were B 5. Because Showthe how tube reduction is so long. m 6m 10 P3. difference 6. lost point between (3).2 of 5 ft/s. 1b2Hg (b)146 vacuum.0.114 turbine of 3.125 static 272 3 in.1b2 2. ■ Figure P5. D y = determine 6 in the power 5. engine a thermometer.78 9170estimate lb   2 of2a.119.122   Q Compilado   H ejercicios  de  Energía:  Cenguel.00of sure the tube has such a small diameter.28 4:00 PM 1260Page minwater 272 is also measured. Chapter 5 ■ Finite Control Volume 1.137 5.118 to0.   No 5. 5. atmospheres.58 !28.122   1. Suppo head loss is 13 ft/1000 5.80 78.5 ft.  White.106) at the rate 17. 5.   Pump 0.58 % 10!3 m3!s pipe 4.the If water allinlosses Fig. If%head 10 W 5. 6m 0 0.104 from sec.8V 2 /2.qxd 2/17/12 4:00 PM Page be changed. frictionless the 4. a 5. P5.112 m!s 12 ft the channel.0125 5.26 1a2 c # " C. !2y. as shown in Fig. P5. If associated with the will differential this flow? reading If1a2 this in be same the 5. where hp is in meters and Q is in 1000 lbm/s.120 theWhat at a is maximum 1610 rate ft the & of lb!slug. The temperature of the 6 m/s shock absorber to move toward each other 12 in. tween and B is500.2)  TURBINAS   pressure rise information.44 2.02 ft!s ■ Figure 5.07 mtruck shown in Fig. r0 ! ri sure lostisbe Gas 6.115 4.104.137.108 4. lb. is arranged nar.28 shock % absorber 10!4 slug!ft stopwatch and shown 3 in Fig. 3 ft h /m .92 !36. and ■ cylinder Figure 1b2 walls. 5.hAccording to   P the pump manufacturer. ■1b2 Figure 3. Neglect gravity and Turbine any friction force between 6.72 0.135 TheApump-turbine m system in Fig.109lbgraduated m !ft An 3 automobile cylinder.112 1 in.42Analysis 352 lb termine Support elevation with elevation h.121.8 ft & lb!slug inlet gas speed is negligibly small.116 P5.48 the average 0 N velocity Oil p = 2 atm 6. bottom 6.ble.8 develops u # ! 100 x hp " " plies to thePump-oil if head losses are negligible. How5.22 psi. ! 160 kPa. will work andbest a stopwatch. *P3.5 Pump 0.105. thepsipump The ■ Fig 5. 150 output m!s 375 m!s heart. where 22.32 720 m!s .36 ft lem if mate the A 5.30 19.110 causes the two ends of the graduated &s cylinder. amou 8 mill 185 r tainer as indicated in Fig.88) a the head flows steady of 600 inrate an ft. the tankflow If the 12 in. laminar What ! primarily (32"LV)/(5. 5. asP3. is 4 ft. 6 that the laminar head loss 1a2 3.105 horsepower which mustabeconstantdelivered to thediameter oil by 1.32 ft.70 is 75 13. Will the actual 4.loss 10 ftfrom below (1) the to (2) whereinlet.113   a 5. where0.2 deg. At night.29 m 3 in. lb Determine If the the turbine head loss develops from (2) 2500 to the hp.52 3.05non-uniform 0.62.qxd 2/17/12 5. 5. w pipe and discharges it through a nozzle. P5.2 V 2/2g. a long vertical capillary tube. 2.08 the valve. 1020 ft!s " 4Q .025) from a submerged 5.40 7.116 P3. (a)pipe Determine has a 4-ft theinside power that diameter.6 N      5.56. xturbin 5. # 2c2y 1x yexpansion is 2 " y22 but with frictional adiabatic 5. When the valve to a2 pressure of 5 psia.18 of flow in the150 liters!s piping. estimate x the power output of a human heart.99 ft!s a very plies simple device to measure the viscosity of water as a 5.118 must Turbine 1. the 1m friction head 5. the 3rate of flow through4.7 lb place pumping stations every   10 mi along the pipe. ■ Figu hp.-diameter outlet of the hydrant 6 m/s is 102. ft 3 diameter !s 6.118a. 50 given 148 ft & lb!slug. 674 when or less. P5. P5.50 the ax #271 150 1x2 mine 2c xBtu/lbm. P5.119 3 !s 5. D = 2 in 5. Show that by varying the length of 4. 2 2 Pumpft!s2 1 tion is the occurrence of cavitation in the siphon. upstreamQ.   point inside-diameter  5.103 24 ■ in. 1440 m!s 2 . ! (1) higher one the same flowrate. The head loss is known to be equal to 4V /2g and the pump less? hrough the P3.56 525 $F!s For the 180! elbow and nozzle flow shown in Fig.132 2.112 6 Section 5. velocity at (3) is V3 ! 2 ft/s. where V is the velocity 5. gal).10 ft2 1m 5.112 What is Water is the pumped maximum frompossible theP5.4 h ft!s p is in ft Problemsȟ1 " when 5. 1c2 . P5.18 Second one P3. 4.10 the hydrant.50 482 N in downward p p   and2 the flowrate is5.58 surize   lake toto thethe oil ifSection headatlosses are negligible.120 (SG across !at0.each pump.plane. 5.46 8860 m!s 2 . withatathis speed 6.16 No P5.00456 Sectionare slug!sknown.837. P5.112.30 14. lami- water siphon having inside 5.124the0. 7440 droelectric m!s 2 turbine shown in Fig. rather if thethan turbine were removed.minimized. 0.113 GO Oil (SG ! 0.75 m 6m 10 2 hp = 20–2000Q 1.The pressure the Determine justflowrate.118 Water is pumped from the tank shown 5. ■ Figure P5. The limiting condi.4 ■ FigureinP5. 1b2 W10 2 " 14.5 line is The 1 in.24and in.118b: p ! 20 " 2000 Q .124 Water level 5 ft3/s as shown in Fig.78 kW or less ■ Figure friction P5.64 3580 lb ■ ■ Figure Figure P5.11 3 turbine 5.24 vu # !4ru ! 9r 2 c lb d 2 in available hf ! cQ system isenergy . 4. The head loss 5.6 ton gal!min. kg!s consists of a tank.0189P3. p1 # p2 ! 75 kPa. v # " f 1x2 free surface is occurs 20 m above the pump.10 P5. 1c2 6 in. The viscometer.116 A pump is to move water from a lake into a large.121 2   hp.7 1-in.128 1a2 0. The Hg vacuum. una bomba.117 ne la turbine útilpd dam. lb!lb am deter 1m 5.18 150 liters!s la pé teachers and educators5.If the lawater jet rises latopre- cual aumenta a day. p1 #rise information. pressure siat la section (2). 5-78   Compilado  ejercicios  de  Energía:   Cenguel. which surized the wa cie del agua está abierta a la includesatmósfera.58 !28.34 !33.58 of wa turbine 5.124 0. irreversible pérdida 10 ft below the de turbine carga delinlet. mate the water temperature change in °F caused by this drop. H hpump. P5. dura Suppor Section (1) height of 27 m from the ground.123. 5. that the pumpIn sup. differential reading L =0 (to get the minimum value for required p in the level. Determine the electric Answers power to Selected Even-Numbered H etry and variable river flow Q.139. com0 turbina 6 m/s 5. determine el gasto volumétrico 20 ft máximo de agua. ay # más 2c2y bajos 1x2 "con y22 el fin de alentar5.131 When the pump in Fig.turbine What=is0. turbine5-70 through a 3-ft Vuelva Estimate the ainside-diameter considerar pump inlet kW pipe power elinproblema as indicated 5-69.    5.102 turbine 0. P3. 0. esti- 0.866 5. ft!sis2 toUtilities be are also 5.130 delivers 25 kW of power 20 m pump.116 When the pump FIGURA   in Fig.113   5.08 hp.128 1a2 0. un tiempo To takecorto 5.130 pipe genew 3 ft ownstream Section (2) 27 m across P3. Show that.52 3. vacuum 27 m bom . 4.0 m !s for precio each alto.05. un empresario considera cons- 5. If the turbine develops 2500 hp. 0.1 $0. genera 0 Tom 5.120. determine 2 the minimum bine–g D1 = 3 ft sión del agua.29M The 5.052 m 3 !s water term 5. gasto is 4 m Turbine h P hpump.2 deg.ρgdirecti sión suministrado por la bomba a la línea de agua.130 P5.3)  BOMBAS   1 m 223del reductor.108 926 ft & lb!slug. for a given penstock geom. When the4.54 !225 ft!stencia .92 power generación !36.   Therefore. is negligible.qxd 2/15/12 9:01 PM Page ANS-3 pump.dores2 parties at 1. u ■ Figu to the water.133 is filled with water at4. andlautility salida 5.133 The long pipe in Fig.117 BMAns.36 ft diame 5.134 1a2 .112 What is the maximum possible powerChapter nivel del lago.25 m3/s.the 78(maximum 3 5.2 g mínimo de pre. delivered the in toDetermineFig. 2/22/06 6:29antes AM después Assumptions A. que 5. (g   5.118 0. 10 ft!s . en la5. Th hrough the P3.qxd 470 kPa y 440 kPa.129 Multnomah Falls in the Columbia River Gorge has a sheer Ing. 132 ft !s losses are negligibly small.4 3. shown el extremo Ifdethedicha manguera se apunta lost between sections (1) and (2).042 head m3!s is 5–82E availa opportunity. respectiva- 1 The flow is steady and incom P5-69draws 220 m /h of water at mente. 5. 1b2 11.this 1c2 22.68 valve 1b2 2 V0 hb!3 must add to theeléctrica water.104 waterprivadas 4.140 148 ft & lb!slug. result obtained The flow rate of 0.  cen72367_ch05.8defor ft & lb!slug po- 2 Suppose a 2 utility company 2 electric 4.64 20°C. boquilla 5 ft /s asque 3 está inenFig.03 dólar/kWh. 1. P3.66 slugs!s Discussion The power 5. u = 27 − 20 = 7 m supo p2 = 10-in. consumi- 4. system moto during ust elimi.96 take plantas $0.08 3.113 GO Oil (SG ! 0.111 open andBased wateronflows   500 flowrate at andmpressure 3 /h.92 a los 1a2 43 deg. pipe y we of at a rate la ■ take the ref drant.132 2. la !8.100 durante advantage 0. Estimate (a) the   ingconspo exit velocity Ve and (b) the   flow rate Q.117 The Water is suppliedthrough flow discharges 3 at 150 fta /snozzle and 60topsi thetoatmosphere.10 314 m!s 5.837. Hg Oil pump– ver condi. hasta 5. 4 yPage Q2 ! 100 ■ Figure P3.117. Se conecta 5.diame © 2006 atmó 4.25 5–83 amount of power output possible? Why will 5.115 2.  P3.-diameter outlet W line 5. con frecuencia. Si la eficiencia mecánica e de la bomba es de 82 por ciento. riodos !25.88) una manguera V2 = flows al fondo 0 andinP1an= inclined del P2 = Patm tanque . For turbulent pipe flow (Chap.16 1. 20°C from the reservoir.   ft 5. 1000 bombearkg/m )(0. sistema de tube.32 720 m!s2. laswill- com- Q H 4.2 20 m!s less? 40 m 5.110 noche. P5.p Q = 150 ft3/s del tanque the a la pressure boquilla estáof water.atturbina.7 kN/m = 68.126. 1a2 4.136 for course1a2 3.5 con- kN 4. Using the steady-flow energy equation.con 1440 5–80 un m!s The . 39. is 10-in.2 Application of the Energy Equation— 3 Suponga que una compañía generadora de servicio of turbine the hydrant público up.116 the lake to directo hacia arriba. P5.20 m!s .114 m abovedurante the lakeellevel.76 much 1b2 234 lb higher su turbogene- 2 Section 4.a razón anddethe heigh . Air 30 m 5-76 Se usa una bomba de 7 hp (potencia en la flecha) para subir agua hasta una altura de 15 m.25 m!s for m 2 3/s electric y la power is usually eficiencia total de5.46 8860 m!s .106 Wlo.32 ft.112? the ce pero 5.7 50 m Lake 5.99 ft!s Depósito Pump– 5. The turbine discharge pipe has a 4-ft inside to de agua y la diferencia de presión de uno a otro lado de ladiameter. 1b2 53.  truir 8 million gal/min across a head of 600W un depósito grande 40 m arriba delft.00 dólar/kWh m 3 !s por la turbine potencia producida durante el %día. P3. 92 alta durante el día during que2 enpeak la noche y. P3.114 5.81 m irr ■ Figure P5.  5.60 1a2 determined from A hos 3 ! plant is to P3. día por empresas pumping 5.5 ft /s to a maximum elevation of 60 ftThe above the hydrant.138 11.qxd 2/17/12 4:00 PM Page 271 25-80avoid La building demanda 2 new de expensive energíapower eléctrica suele plants thatser mucho will be más used Penstock 4. Discussion and its overall turbine–generator effi-aboveHg repre vac properties of water.qxd Waldo  Lizcano.       Fluye W hite. Para aprovechar esta oportunidad.much 375 m!s lower prices to encourage 5. Tomeis open to the and its di.122 1a2 4.112 density of water to P5. The river D2 = 4 ft 5–81 ingre 12 in.6 1.120output 1610offta&hydraulic lb!slug. A 20 m water column height of 7 m corresponds to a he atmos. determine la pérdida irreversible de carga kPa. Com.3.00 m3!s PROPRIETARY MATERIAL 40 m 5.02 ft!s sumers 60 ft to use the available 5-79 power generation capacity 5.113. El tanque low are 1 ≅ a (Vestá 0).   en M unson. u = z 2 − z1 la pé efficie Water 4 m 10 ft Substituting.17. 6).  P3. mercury and the water manometer surface is 3 ft.62 5.78 companies de 9170 potencia often lb eléc- 2 2 2 4.122.1 pico. determine the power rías es de 5 m.14 0.132. the friction head loss is 4 m. 1b2 20 a short ft!s 10 2time psi.80 78.7 kPa Answer P3.57 in. 7440 m!s Solution The available 1c2 head116ofNa &hydrau m. u = α2 +inver 6m ρg 6 in.0189 4kg!s  Se  r3ecomienda:    5-­‐78. pe-right to le oflosthis 4. pressure at the 4-in.116amount the actual Will work be forapun 2a droelectric turbine shown in Fig. en la línea central es de 1 ! 9 cm.42 8.734 y desea ft   sure 5-82 is †5.58the from a%un10!3 ■ mFigu level a 2°C. as in Fig. determine el+ aumento pressure rise supplied by the pump to the water line.20 3. Si el chorro de aguaP1se elevaV1hasta una altura de P2lem moto if t 27 m por arriba del suelo. drop of 543 ft.4-in. 5660 ft & lb!slug If head= η turbine ghturbine ρ Section 5.72 3000 kg!s.5. the pump must supply a minimum P5. D = 12 cm De = 5 cm hP5. shows α1 + z1 + hpump.  Respuesta: 0.$C!s Estas compañías también Analysis Whendeseando están the 5.12 !0. the maximum turbine power this turbine. where 2 5. the   total friction head loss is 5 m.63 ft!s FIGURE P5–80 Bomba. pagar1b2 0. p2 ! 160 estimate kPa.119.4 turbine periods.90 determined. what   4.035 m3/s.4 $C!s.   duce de 15 cm hasta2:25 10/29/04 8water cm is PMmediante Page   223un reductor.     5600 . determine the power5.88 1a2 231 and to N & m.22 0.48 1a2 10 ft!s2only . larger pressure rise will needoutput to be of supplied lostto beto ciency is 78 percent. is open calculate to the the poweratmosphere.132   2003 $C!s. ELEVACIÓN  VERTICAL   possible in this case is Pmax ! 2"gHQ/3 and occurs when 5-79 Una 2 turbina hidráulica tiene 85MAXIMA   m de carga disponible the flow rate is Q ! !H "/("3 "C ")".   50 m Pump 5-77 agua unSolution tubo horizontal.115 The Reservoir hydroelectric shown in 5. a 0.8 ft!srador during.0125 lb5. a hydraulic  los factores de corrección de la energía 6 m/s cinética como 1. 2 presi FIGURE P5–78 = 68 . 185 c05FiniteControlVolumeAnalysis. P3. Hydrant an entrepreneur is considering building Problems acomprar 271 large 4. and tem.50 ax # 2c2x noche 1x2 " ay2precios 2.36 1225x " 1502 sell m!s trica power de esta .33 night construcción and ft!s Properties is willing de nuevas toWe pay 5. it is atDetermine night.106 = η turbine is 10mpsi. p1 # p2 ! 75 kPa.22passes rise S 90 the power output of a human heart. Assumptions 1 The flow is steady and incom que usen la capacidad disponible is selling de 5.711.03/kWh y evitar ft!sat . !28.48 am!s diameter a2 high price. 150 night m!s at . drains 5. Substituting.demand gasto 2160 de 0. el gas- water.6 N & m!kg.130. A cuyo hose diámetro connectedse tore-the botto C through   P3. la reservoir ■ Figure energía100 $C!s 40 producida P5. suelo.6 hp pañías ing to generadoras purchase power de servicio produced público duringvenden the daylafrom energía en la private 4.130 3 Cengel 05.43 ft & by preparation.nivel the del energymar equation y la superfi- for steady i Ve agua 2m Pump pliesleading line to the oilfrom if head thelosses tank to negligible.78 kW 5-83                                                                                                                 factor arrib 5.132 Consider a turbine extracting energy from a penstock in a ■ Figure P5.56 525 $F!s costosas power produced que sólo during se utilizarán the day. 3 The waterensurface éste.05 esthe de day 78 por thanciento.112 Fig.68 m Turbine Properties We■ take Figure the P5. 0.48 hp so y output of5the hy.8 4.08/kWh !3130 thegeneradoras density forofftwater & lb!slu to Turbine 4. pressurized by a pump. 0. the nozzle is a pump.56.114 is to deliver static sión. the pumpEn la and línea the que waterconduce stream reduce bara p1 = 60 psi increases or less. 2 With Shaft Work   vende energía that the pump surfac affect the When valve A is closed.126 1. The 5-78 El nivel del agua en Analysis un tanque We take pointarriba está 20 m del free surf 1 at the to deliver Figu static bomba. Also. 5–79 3 A hydraulic turbine has 85 m of head available at a p the constant C depends upon penstock dimensions and the 1. the 1b2 sy 9. Si se mide pump que la ispresión to be determined.   5.114 The pumper FIGURA truck P5-78 shown in Fig.117 the friction head loss is approximately hf ! CQ .791.     5–84 des a m !ft 5. min Δ P = ρ gh = (1000 kg/m 3 )( 9the5-81 .24 1a2 6280 kg!s . Hg 5.ofP3.119to Selected Even-Numbered Compilado   Homework ejercicios   Problems ANS-3 de  Energía:   Cenguel. ex. 3.1b2 2004îN!& 4ĵ m.120 data from (a) 22 an ft/s.29 m! .123 ■ Figure P5.90 turbine 39.156 (a) 102 kPa.148 dischargeshand f ! 0. If the turbine develops 2500 hp. N !2y. 1c2 0 head is tion 20 " hp !rate. The river condi. rm 6. m 5.80 the 3.92 1a2 43 deg.58 !Ar 2 4.48 1a2 10 ft!s 2 .106 D = 12 cm De =' cm ! 75 rad/s 5final 2 3. Heat-transfer. esti-! g 0 m!s2 1b2 234 5. mental data.8 ft & lb!slug Ve on— pressure at the 4-in.7 kg/m . 558 N & m!s BMAns.92 (1) dV/dt to ! gh/(L # h) 5.118 1.  White. 75 kPa.7 lbThe 5.120 Water5.ft160's .90 39.36 1225x " 1502 m!s 0.140 148 ft &   lb!slug. what horsepower is needed to drive 4. (Q0200 ft & lb!slug 23 6.113 ft ) lbf. Suppose that this pump 5.116 A pump is to move water from a lake intoplant a large.106 5660 ft &Why output possible? lb!slugwill the actual amount be these variables.96 loss is2 5 m.152 (a) 85.66Nft!s # 116 & m.112 2.94 ftdevelops 100 hp Determine the aerodynamic 1b2P5.4 ft & lb!s Problems 271 3.102 h2/h1 ! %&12& #&12&[1 # 8V lake to2the higher one at the same flowrate.loss is known 2to be 1.p ! ng.90 available dV/dt effects mate the water 3 temperature change in °F caused by this drop.6 kp/[(k develop? hp $] # V 2/2 # 6.76perature lbare negligible.110 734 ft 0.07withm3the/s ■ Figure P5. 375 m!s 5.4° P5. P3. 1c2 22.EES 6 10 ft 5.116 Will work one inlet forpipe not4.132 of the Consider pump lack of the pipe a turbine importance) diameter extracting of kinetic energy 90from energy mm.126 1.22 % 10 W P3.1 in.P3.116 Will work for 2 atm.115 passes steady 3.12 !0. 1110 ft & lb!s   a 24-in.ft!s 2 does the 5.154 0. (b) Here are some actual3. v 0. !28.24 5. 4. 1c2 116 N & m. If heat losses to the atmos.65.30 /s 1a2 0.58 3 is 20 ft /s.126 60 psi 1.132c2.9 c 0 m!s .796 # 21x " y2 " C.  W aldo   5.112 2. resulting per.125 A pump exit above (compare results with and without inclusion moves waterVhorizontally velocity and (b) the at flow a rate rate of Q.130 ! 61VWhen 2 2 ft2the s2pump .Fig.32 # !ft.108 (a) V ! V0 /(1 # CV0t 5.78 kW formance curve as in Fig. estimate the amount of Section (2) ft!s2.78 Fig. 1b2 53. atm. 92. gpm. 50 ft !sflowis the maximum m What 1 Section (1) Pump an expression 6for m the viscosity of the water 3.134 error 4500in thehpcal- !s by Compute the percentage Head.26 1a2 c # " C. P3.. 1c2 22. 0. 2 Chapter Q 3. P5.122   ■Answers Figure P5.94 water h ! 0 at t ! 70 s m!s .134 1a2 17.116 lb at a rate of 1000 gal in 10 min 3. Q . .6 mto/hin-   600 ft river conditions (Q0.7 m deep.. 1b2 c # !A ln r " C (a) D = 6 in (b) 5. 132 ft !s 5.85. pres.4. inside-diameter piping reservoir. estimate 5.120 10 1610 ft below the turbine 2xft & lb!slug.33 ft!s du/dt ft 0 /L)(1 # lb!slug ■ Figure P5.128 Q 6. 5.2p2 !V!3130is the 160 (a) kPa. aThe P3.2 right to!leftdT/dt 125 !s 5. sure is 120 kPa.43 5. (c) Compare the experimental 3. " ! 998.5 ft /s being pumped from sections 3.104 4. 3 ft Q V Pump Pump 202 Chapter 3 Integral Relations for a Control Volume 808 Answers to Selected Pro Pipe area = 0.22 % 106 W (b) Det 6.36 ft   P3. 185 rad 2 3 3 left 2 2 the pump head and the flowrate is3. d ! 0. Chapter 6 p2 = 10-in. 20 ft P3.160 (a) 31 m3/s.127.axthrough P3. The loss of3. 375orm!s less.32 " 15z'720 ! 1y2'. 3.8 4.29 m!s.142 typical a head for4V /2g and a given the pump shaft rota. Downstream dam. 0.120 5.120 pthe 2 !kN 225 kPa. 5.4 kPa.rad!s.106 56600 ft & lb!slug 6.130 34 kW 2 is 450.122 to deliver static pressure at section 6. where /h.   1b2 11. f 0. isExplain 3.102For the0.-diameterculation stream of of $ airwhich havingwould a uniformoccurspeed 3.72 1a2 3000 kg!s.146(Chap.2   deg.asQ2 ! D2 another (2)100is lossP3.6 N & m!kg. Hg vacuum. gravity. the pump Forwhere turbulent the pipe pipeflowdiameter 6).08 ft a 1c2 5.120 phere and ground are negligible.122 2 1a2 4.118a. Repeat a# the prob. 4. energy associated with 2. amount of loss is associatedrection.24 m!s.153 3.90 1a239.110 734 ft age velocity in the pipe. r " 4.43 ft & lb!lbm.56.5 kW experi.54 y 0. (b) 2 8. and report a per.58 head%of10 !3 600 ft.182 meters and1c2 116 are N negli- Q% & m.170 cm h ! 1. 5.041 in. p2 !P3. Vrad!s.144 675 N 5. 0.042conditions 4.4 flowrate. (b) 1 atm 5. shaft The flow discharges through a nozzle to the atmosphere.5 1a2 231 as lbm/s. 1a24xt 231.5 psi ft & lb!lbm. 400 kPa surized tank as shown 5.7 lb in afriction problemhead suchloss as this.!1b2 9r9. Section (1) D1 = 3 ft 5.32 ft.22 (2). According to the pump manufacturer.aver- !& (2V 2 !63.2 with appropriate 2 calculations. Determine 10 how ft!s the 2much horsepower .42 8. 1b2 9.5atm.5.141.106 5660 ft & lb!slug 5.158 (b) 209 60 m!s 2 5.122 L ! %h (cot ()/2 periment: Water T ! 16. Tdeg.112 2.5 is 1) in hra # !rm0v!s 4.32 ft. de. If head 8-in. turbine at 350 lbf/in absolute. 3. 20°C from the the 8-in.5. 100 $C!s z 2 = 150 ft 5.36 ft vacuum 3 s 6.  Munson. P3.-diameter outlet of the hydrant is 10 psi.139 theTheconstant horizontal pump in Fig.140 Steam enters a horizontal from the tank shown in 5.100u 0.138 factor of-power in ! part gd4(H # " +$(b) L)/(1 5.114 (a) 414 r/min. andWhat p3 ! is 265 thekPa.88 pute 1a2 energy the rate of associated 231 N shaft m.amount of power 5.178 78. P3.100 507 m/s and 1393 water.1 # c of the hy. (b) 88 m t a rate of 5. 200 ft & lb!slug 3 3. viscosity of the water in kg/(m # s) based3. work &with this done 185 flow? rad!s.76 1b2 234 lb 3.48 m!s 2When valve A is closed.0 deg.172 1b2D234 ! 0.122 4. must elimi.76 m T + ∆T 0 ft!s2.52 value of3. (work done 5.110 (a) 0. steady rateD3of ! 80 4 cm.0 m 3 !s for each line hp = 20–2000Q2 P5.132lb ft 2c2y 1x2 " y22 5. in Com.140 (a) the 1640 hp 1   when the flowrate5.78 kW 2 ■ Figure P5.96 #4.174 (a) 5. 1b2 53.112 properties water.2 D22=.48 55 MW hp of waste heat to the river.131 When the pump in Fig.6 (a) 6V /L. The turbine5.36 # 17.10 heat g 5.4 u "558 C.130 delivers 25 clude the kinetic energy flux correction 3. whereinVFig.2 d Z/dt # 2gZ/L ! 0 loss in available 5. 375 580°C. drop of 543 ft.14 0.168 h ! 1. !8.042 The hydroelectric turbine m3!sshownIsothermal in Fig. forIfthis 1b2 this 200device N &and same m.12 plantNo heat exchanger? How will the4. (b) furthe 2 4.118b: heat P3. 111 5.128 1a2 0.108 926 ft & lb!slug.at78.P5. p1 in#Fig.115   T Homework Problems ANS-3 etry and variable river flow Q.132 losses are 7 Btu/lb 2 of steam. atm. 1b2 3. The river published value of $ at this temperature. 2determine4.5 rad!s 3. rslugs!s m!s # A!B 5.122 6.8 losses ft & lb!slugP3.116 5. P3. 200 $C!s. Water is pumped from the large tank shown in Fig. to the water.1285.114 5.92 z !36. 3. the relationship h ! 2.136 1a2 3. 1b2 11.164 (a) 5.. and (c) the flowrate expected if the turbine were removed. (b) 2.796 Q possible in this case is P max ! 2 " gHQ/3 and occurs 120 kPa when (a) 169.40 Yes.78 h.68 the power 0 P5. . m3/s adds to 5.38 1a2 3.0125 lbm !ft3 5.127!3130 3 ft2 & lb!slug x3 Q0.128 1a2 0.141 Water y22.6y ! 1001b2 11.102 0.143 ing in the 5.100 0. If all losses are negligible.136 forgot if a student of kW40 ft/s.134 1a2 17. none 4. equal to 5.  Water is to be moved from one large   reservoir 3.132.88 2x.92 with from water at the 1a2 43 lower 20°C.5 sketched in Fig.120 1610 ft & lb!slug.102 0.4yt 185 # 0. Deter- varies thehflow rate.lake to ! 7 cm.2 # 2xft! 4y " C.6 0.N1b2& m!kg.20Q affect m!s 2 the .116 tionsWill upstream work arefor Qi ! 2 2. 1b2 4.14 downstream 1a2 2.128 result 1.126 1.3.114 nate 5. 0.& Nolb!slug head2.02 ft!s betweensaturated conditions. 3.133 m h ! Page 5. 3 right to left 50 ft 3. L ! 36.deg.117. (b) 68 TankkN 4.48 !28. 1b2 3.xright " to " f 1y2 4. rad!s.114 5.162 Q ! 166 ft3/min.1 ft!s . the friction head loss is 4 m.50 # 2c2xthe1x2 " in m3/s.176 average (a) velocity 9. 2160 m!s 5.126 926 ft & lb!slug. p = 2 atm 1b2 53. 150 ft 19.21 20°C m C depends upon penstock thewa- 5. if the power plant is to m!s . 0)? x 4. Using the steady-flow energy equation.132 2.138 11.144 flux athe pres- penstock correction 26 kWin a factor turbine.58 3 Water is5. and 12 ft/s and is dischargedm!s Turbine at3. 3 5.57 in.134f1a2 6. 1. with pumping the fluid from the lower Estimate the pump power in kW delivered to the(a) 3.133atis1500 filledgal/min 5.124 of water0.6 ft the is approximately hf ! CQ2. 22ĵ m!s 2P3.13660. 1). 1b2 3. (b) termine (a) the elevation 5.166 (a) 60 mi/h. as in Fig. T0). 2 .29 m!s.with where p is in ft when 5.122 vu1a2# !4ru 4.130 is the draws average 220 m3of velocity /h of water at 2 2 3 # # 3.96small.56 525 $F!s At (2.94 f 5.34 5.56.126.96 P3.124 41 r/minthe m. 1c2 22. as in Fig.22 % 106 W cient. P5. !sf 1x2 The 5. If the pump 4. the& m. P3.16 No hp ft3/s and 60 psi to a4.48 hp p1 = 5. Yes. pfrom m by gravity 1 ! 150 one kPa.104 across a4. 150 m!s .(b) 110 ft/s less? 5. Penstock to Selected Even-Numbered Home Answers lem if the tank were pressurized to 3.16 4î !m4ĵ!sft' s2. Q1! 5.03 ft (subcritic 6. the total friction the Determine headamount 3.118 0.1381a211. 92.119 P3. 22k̂.124 0.36 ft & lb!lbm livered c05FiniteControlVolumeAnalysis.121 efficiency of the fan.052 m3!s 5.4 ft!s 5.34 !33.45.6 Turbine 6. (b) L ln 3/(2 6. (b) 54 kW pump sup. shown 9.00 m3Pump !s 5. f # !4x ! 2y " C 20 m!s 5. 5. Pipe Chapter 5.00 5. a #1.132 2.18 Second inside-diameter 5.108 926 ftconditions& lb!slug.P ! $WriteQr2*[r2* % Q c 3 ture is obtained by weighing a known volume 8 million gal/min 5.052 # m" 3 4.133 y #long 2cat2ypipe 20°C1x2in" isFig.116of water. power required. N u#& 3.92 When friction the4valve !36.8 # 3xyAir ftk̂.127 A power plant on a river. 2 where V is5.118 column 3 = 5.3 m/s. .02 m 3 of kinetic /s. and tem.58 % 10 m !s !3 3 !y 2has 4.7 where kPa.118 discharge pipe 3 a 4-ft inside diameter.142. 1a2 forin3Fig.05 Water is pumped P3.P3.61 ft/s.05 734 ft 0.33 ft!s2 5. h ! 3. ft 5. 7440 m!s P5. 1110 ft & lb!s inlet. T0 in the pipe.583 Q isininaftpump /s. rather 2 than 2. 5. of a200 wat drant.60 as indicated 2 atm.08 inhp.5 rad!s 272 Q Chapter 5 ■ Finite Pump Volume Control D = 9 Analysis H13. If head losses hp ! 20 " 2000 Q .92 !36. 2x. Show that.875 5.140 148 ft & lb!slug.94 cosftu " f 1r2 1b2 2 V 0 hb!3 0 5.5 that 3 hp to kN the water work for this purpose? Support your answer 6.104 the #nearby 4. p15. 91702 lb The 6.7 lb ing. atmospheres. Ti energywhere e flux correction factor).12 2 m!s.310 mL/s.10 lost between sections (1) and (2).052 m widemand 3 2.180 shown V ! in Vf tanh (Vf t/2L).8 ft!s . P3.as to the upper reservoir. 2 andNFig.130 centage error. and H ! 0.6 1.142 the (a) 1150(or importance gal/min.78 Q1 !9170 flows 220 lb3/h.139dimensions3.10 314 m!s EES mine the flowrate.875 Power the flow rate is Q ! !H "/( "3"C ")".5°C.62 5. Br cos 5.64 is 10-in. 3. P3. is de-(b) 55. (b) 317 .100it? 0. for a what given power penstock in kW geom- Turbine 3.10 ft2 ■ Figure P5.141 Pump 5.9°.126 There is a steady isothermal flow of water at 20°C through P3.118 P ! $QuV as a function ofn(cot .4 # #!225 ft!s . Q.58 % 10 river !3 water m3!s by no more than 12°C.119. 1b2m20 3 ft!s /s.4 ft 5. !25.0189 kg!s 5.33 ft!s2 losses are negligibly 5.127.of its 160 di. 0.36 ft & lb!lbm P5. where 5.4 ft & lb!s 2 3.78in Fig. inside- 3.88 to another the device in Fig.115 5.88 2.56.4 is 75 percent effi- 76.07 m z1 = 50 ft Q. 5.139 2D2 = 3 cm al!min 1440 m!s . 4. & m. !3130 determine ft & lb!slug the power that the pump 204 diameter Chapter pipe3 Integral 2 m Relations for Pump a Control Volume 3. Estimate 3. what Pump should be the r ! ri minimum flow rate Q. the maximum turbine power 3. estimate the downstream 1610 ft & lb!slug.08 hp.76 3. 2 3. Known data are D1 !at9 cm. The mass flow is5.123.2 A ln deg. (b)kW   pressure differencei across the the UpstreamP3. V #5.6 hp 3.2 deg. 4xt ft 4yt2.80adds678. 2 4Q .042 0m !s ■ Figu hp.104 ' ! (%Ve /R) ln (1 % 2 2 c y 1x " y 2 2 pumping power required. a higher elevation as indicated in Fig.108 P3.118 5. 3 Neglecting losses. 92.145 5. pumped y22 P3.25 kg/s.1 $C!s 5.5 110 ft/s and 25°C kN !8.qxd to the water by the pump? 2/17/12 4:00 PM 272 Answers ■ Figure to Selected Even-Numbered P5.qxd 2/15/12 9:01 PM Page ANS-3 ter at 57 mof/h.80m /h. 2 gible.121 turbine1a2 The5.123 Lizcano. 1b2 hpv # !1y ' 2 " z2î4.       P3.08 hp. 0.110 0. areopen is 2 approximated and water by hfat!500 2flows 27V20m/(2g).2 Will a pump 5.99 ft!s P3.4 ft!s2 6. where hp is in 3. 3 m /snotandforTi !3 18°C.AThe head pump loss is has known to bewhich. involved.129 Multnomah Falls in the Columbia River Gorge has aVsheer ! " # ["2 # 2Vj]1/2 e 271   Ing. 9170 P5.124 A 34-hp motor is required an air ventilating fan to produce 5.126on these%15.66 p.124 0.2 V 2/2g. Calculate 3 6 s 5. 4!2y. lb y is m Chapter 5 5.46 8860 of Prob.4 $C!s.117 supplied at 150 hydraulic 6m 10 turbine through a 3-ft 6.92 No 1a2 43 deg. P5. (d) 3.36 6.112 T ! ṁR 02' must add to the water. 4.124 A 34of Turbine -hpthe pumpis where motor requiredtheby 6. Determine the amount ft loss in available energy associated5.92 the powerftthat !36.132 2. 5.66 % hydrant.34 !33. m 4. If the of the pumpturbine wheredevelops the pipe2500 hp.20 estimate the amount of 20 energy 3.7 ft !s 10 ft Determine the aerodynamic efficiency of the fan.92 (1) 1a2 43 deg. P5.05 amount of5.120 1610 ft & lb!slug.4 ft!s2 from (2) 5.119 required. fromQ. diameter pipe lake to the higher one at the same flowrate. 1.80 375aerator.pump in the Application Section 5. Determine Pump 5.119 13 ft Water (a) is pumped from4.125 A pump moves water horizontally at a rate of 0.07 m oftothe theEnergy pump manufacturer.to4.118 lake into0. P5.114the flowrate when The pumper of watertruck is 20 ftshown 3 in Fig. The head loss is known to be4.   (3). 3 m !s with 2. 1.rad!s h p.123. point (2).791. 1c2 22.104 4. estimate the amount !4■ Figureof 3 P5.88 the 1a2 power 231 Nthat & m. P5. If all losses 5.48 hp of water sketched 5 ft3/s asinshown Fig. 1b2 62.2 pipe.! the9rpres- 2 cos u " f 1r2 turbinekN. associated calculate with Pump this the power 5.26 1a2 the 0.46 1a2 6. 674 lb ■ Figure P5. P5.2 0V 2/2g. 5.32 1a2 15.   velocity at (3) †5.10 ft ■ Figure p = 2 atm 5.34develops 120 N100 hp P5.796 sketched in Fig. where y is m 5.58 % 10!3 m3!s Pipe area = 0. m3/s the relationshippressure adds to the at water the 4-in.63 ft!s ft associated with 2. !63.116 s 6.7Albpump moves water horizontally vacuum 2x of 0.6 N 6.120 at the steady 5. P5. Downstream of the pump where the pipe diameter 2 5.reservoir to another Q Pump Pump 5.1a2 The 3. !28.96 (1) to !3130 (2) where ftV &2 ! lb!slug 0 is 4 ft. where V is the average velocity 3 P5. a # ■ Figure pumping P5.900 lb turbine.140 148 ft & lb!slug.16 and271 m5. to the top droelectric turbine in Fig.5 Video kNjet and Fig.0 92.112? 4. According 0.column Aerator (a) Determine 5.10 ft 2 Aerator column 4.36 1225x in " Fig.120 in Fig. 1b2 20 ft!s2. not for 3 loss mercury Q in available manometer energy is 3 ft.112 2.6 1. ft lb is 120 Section lost between sure (1) (1) and (2). 558 N another amount of loss is associated with Section (1)at a higher elevation 50 pumping ft 5. in available energy associated 3 ft 5.114 5.50 ax # 2c x 1x " y 2.24 m!s.5 psi 5. !2y.16 pipe an No is 90fan diameter air ventilating mm.023xy mk̂3/s.22 vat # Hg " f 1x2 3 5.1 ft!s .5 ft3/s being pumped Sectionfrom (2) sections (1) to 5. 4xt2.116 shaft power Arequired. pumped the tank shown in Fig.18 the 150fluid from the lower liters!s Pump as indicated in Fig. lb 22 (2) gal/min across a head of 600 ft.115 passes hp. 5.0 deg.122 2 point 2 (3).119 6 m/s 4.20 m!s .90 to the39.125 p2 = 10-in.123.10 314 m!s 6 in.6y ! 100 kPa.16 with this 1.2flow? that the20If pump this same m!s sup.00associated energy (3) 3 ft /s being pumped 5. termine (a) the elevation h. 5.68 66. 5.123. 1. the steadyIf therate Chapter of 80 gpm.60 1a2 200 $C!s. is inaddpoint Free jet mP5.88) flows infrom one lake an inclined to at pipe another a rate as of (2) is loss ! 61V 2 # 2 ft2 #s2.4 deg.of 5. Will ■ Figure P5.33 (3)(2) ft!s2 Pump 0.119 Water shown is pumped from the large 20 hp = 20–2000Q tank shown Water isWhy pumped willfrom the actual a tank. /s.24with this 1a2 6280 flow?kg!s If this same of shaft power required.05 0.0 m !s for each 3 ■ Figure line P5.123 Fig.29 m!s.64 132 tankft3!sshown in Fig.118b: 2 h Answers to Selected Even-Numbered Homework Pro     is V3 ! 2 ft/s. tothe pres- produce 5.118 2 3 p ! 20 " 4Q . differential 5What in reading is the the in the 8-in.02 ft3/s.69 lb Q =3 150 ft3/s Section (1)5.5 is 4 ft.121 60 ft 3 76.34 c # 2x ! 4y " C.875 12loss in. 5. 1.hydraulic ■ Figure P5.0 ft3/s.6 m Pump 0.119P5.-diameter stream airPump ofinside- 8-in. The head loss is known to be equal to 4V /2g and the pump head is hp ! 20 " 4Q2.3050 ft14.07 m !225 Q . none when the flowrate ■ Figure P5. P P5.121 turbine shown in   Fig. 3120 lb!ft !Ar Hydrant Section (2) 6.were45. ■ Figure P5.6 N & m!kg. pointPump (3). where hp is in ft when Q is in ft /s.42 352 lb p1 = 60 psi 6.10 2 50 m 4. 1b2 Yes.74 1a2 181 lb. determine diameter is 90 mm.121 4.3.. if the average 2 Water is pumped velocity from aattank.10 g # 5. de.   The hydroelectric ■ Figure 2P5.54 28.05 0. No 5.32 10 720 m!s2. a rate termine (a) the elevation h.520 3. Support your answer with appropriate 5 ft calculations. where V5.118 /s.042 (3) is Vm 3! 3 2 ft/s. 2.36 ft Oillake to the higher one at the same flowrate. 5. water plant aerator. as shown in Video V5. 0.12 one lake tokg!s 1a2 0.117. 1b2 c # ! 5.ft!sDeter- 0.60 6.119 in Fig.5 3.0 ft3/s.54 ft!s 3 . 9.122 power output possible? 5. !s Q ■ Figure head V is hPump P5.8 kN ft turbine. (a) Determine 5.-diameter stream of air 6. 6. 1b2 ft 3 /s.118a.0125 80 gpm.(b) ft!s10 10Determine 2 ft the head loss .01 ft3!s. P5. 1b2 9.2 one large reservoirdeg.43 loss ft oflb!lb & available m.08 hp..0 ft /s. estimate 5. 1c2 675 N a 24-in. and (c) the flowrate expected if the turbine 5. inside- amount of loss is associated with pumping 5.124 A 4 -hp motor is required by an air ventilating fan to r 0 ! ri produce D1 = 3 ft 5. if theW water.5 ft3 /s to a maximum elevation of 60 ft 5.123 tank were Water is toto pressurized be3.7 61V 2 # 2 ft2 #s2. 8-in.64 3580 lb 2 ■ Figure P5.118 3 /s.32 bottom ft. Determine the amount 0.52 17. pump is to move water from a 5.the 3pres- thepower x3 turbine.50 482 N downward negligible. TheTurbine head loss is known to be 4. where hp is4.57 in.121 5. The turbine discharge pipe has a 4-ft inside diameter.8 m!s 5.121 diameter The turbine shown in5. where V is the average velocity of water required.0189 2 Pipe area = 0. pressureQ.40 7.1502 m!s less?2 . 2160 m!sAerator column12 in.120 power pumping Water flows by gravity from one lake to another as (2) is loss ! 61V 2# 2 ft2#s2.56 the large 525tank $F!s ( b )shown in Fig. Downstream 24-in. H 5.02 lbdevelops 100 hp when the flowrate of water is 20 ft3/s. as5. inside-diameter lbmWhat 5.58 atmospheres. 1b2de-W2 " 14.02 m /s.128 diameter pipe lb from !0.113. is 10-in.711.68equal 1b2 to 4V 2 2 V /2g and hb!3 the 5. Ing.62 1290 lb!ft. !p4 0Ifishead 4 ft. of 80 gpm. (b) the pressure difference across the Upstream 10 !4 The m !s Hg vacuum. u 5.123 pump Water is to be moved from one large 5. as shown inProblems of aThe Fig.00flow?m3!sIf this same of 5. f # 21x " y2 " . large 0.  M average unson. 5. . 1b2 11.66amount the slugs!s of or less.26 1a2 c # " C.106 5660 ft & lb!slug ■ Figure 5.14 of the1a2fan.-diameter if the head loss outlet fromof(1) thepto3hydrant (2) where is 10V2 psi. Pipe area = 0.amount point (1). 1110 ft & lb! a pump that adds 3 hp to the water Section work (2) for this purpose? pumping ■ Figure power P5. 5. 5.16 2 P5.46 8860Qm!s .123 # !r 0v ■ Figure P5.44 2. 10 psi and (c) the flowrate expected if the turbine 3were removed.28 % 10 ■slug!ft Figure &P5. Downstream ofPump the pump where 6.112 What is the maximum possible power 4. 14408m!s million .866 in the 8-in.120 5. termine (a) the elevation h.052 a(1) large. V # 0.2 2x. What 0.28 1260 min 50 ft 5.125 at6. at a higher elevation as indicated in Fig. 1b2 392 N. Deter- mine the flowrate. 10 ft below the50 horizontally turbine fta rate inlet. and (c) the flowrate expected if the!185 removed. sure is 120 kPa. Determine the aerodynamic efficiency 6. x # 5. (b) the pressure difference across the Upstream 5.120 power required. of the fan.4 60 psi 1a2 to a0.119.124 A 34-hp motor is required by an air a # 5.  Waldo   5. 4yt2. having a uniform speed of 1b2 v # !1y'2 " z2î " 1 40 ft/s.66 2.22rate 0.115 5 ft 5. hite. 5.108 926 Water is pumped from the large (1) ft & lb!slug.8 & lb!slug the pump 0 0. where hp is in ft when 4.8 output of ft!s the hy.5 ft3/s being pumped from sections (1) to 5.120 5.119.113 a 24-in.114velocity 5. Deter.4 the topft & lb!s ■ Figure P5. (a) Determine the power 12 in. the185pumprad!s.134 1a2 17. 100 $C!s 5. hp ! 20 " 2000 Q . 1b2P5.8 lb. 1b2 2 2 of a water plant aerator.122 (3) 5.12 having a No uniform speed of 40 ft/s.120 at the steady 5. inside-diameter piping involved.18 pump a where Second uniform the speedpipeofdiameter one 40 ft/s.116 at a rate of 1000 gal in 10 min are negligible.4 and the pump$C!s !25. if the average velocity 5.6 gal!min. !ft3 is the in the 8-in. point (2). 150 m!s2. (b) Determine 3 ft the head loss from (2) 5. 13 ft 5.07 m 10 (a) 4.122Repeat 1a2 4. head loss is known to be 1. 7440 m!s2 adds ■ Figure P5.!y 2 5.02 m3/s. 1b2 20 4.36 50Airftpiping involved. 1b2 146 lb 6.102 at 0. If all losses are 5. !8. 5.120 h Water flows by gravity another as (2) is loss !P5.124 one 2.70 13. 200 ft & lb!slu P5.4 turbine Determine through the a 3-ft inside-diameter gal!min aerodynamic diameter pipe efficiency inlet pipe as indicated in Fig.22 % (1) from sections 106toW mine the flowrate.126 P5. rise information.6 bottom hp of the aerator p 2 2 2 column.138 11.136 P5.30 19. V atP5. of a water shown in FreeV5. Video V5.plant m!s2 point (2).76 T 234 in.96 lb ■ Figure P5.10 P5.121 The turbine shown in5.110 reservoir 734 toftanother ■ Figure P5.78 9170 lb be 5.8 the upipe ! x 2 " " f 1y2 # diameter 2 3 5. What is the 12 maximum 5.58 equal to 4V2/2g$C!s.48 m!s P5.48 h1a2= 10 (bft!s ) 20–2000Q2 2 . right of thetoaeratorleft P5.2The 5.2 ft/s.117 Section (1) Water is supplied at 150 ft /s and6. The loss of available head is hp ! 20 " 4Q2.36 above the 2.122 (2) 10 ft 5. m3/s2.112 0 at a higher elevation asTurbine indicated in Fig. (3) is V3 ! point (1). Equation— Q. m mine the flowrate.1 column.00456 fluid fromslug!s the lower increasing.-diameter stream of of having air the6.117 6.99 ft!s lem if the 5.115 L 2 izcano.118 4. P5.05 2 0. P5.8 4. is the0.122 is the average velocity of water ■ Figure sketched P5. ay # 2c y 1x " y 2 5.16 and Fig.48 0 N develops 100 hp ■ Figure P5.38 60. P5. 1m 4.119.Pump inside-diameter piping involved. (b) the pressure difference across the Upstream of the pump where D2 = 4the pipe diameter 5.837.121 to the water if the head loss from (1) to (2) 1c2 where 116 NV&2 ! m.116 Willtheworkamount for 2 atm.40 1a2 Yes. No 3static pressure A pumpatmovessectionwater (2). mpres- 3 !s amount plies to of theloss oil ifis head associated losseswith pumping the fluid from the lower surized tank as shown in Fig. Pump between the pump head and the flowrate 20 is as shown in Fig.66 ft!s2 3 ventilating fan to produce 5.6 zof#0.122 at a rate of 3.10 ft2 lake to the higher one at the same flowrate.moved rather from than 5.56. 1b2 3.122 at a rate of 3.122 at a rate of 3.113 GO Oil Water (SGflows ftby gravity !130. 0. 5. on Determine Compilado  ejercicios  de  Emust nergía:   toCtheenguel.100 to the0.3 lb 6. estimate   the power output6 mof a human heart. sure a is 120 kPa. where hp is in meters and Qcolumn.4 3.14 rate of 0.111 Based flowratethe andflowrate. !28. f # !4x ! 5. m /s adds 600 ft to the water if the head 5 ft loss from 5. sections kPa. 5.5removed.123 shaft power Water is to be moved from required. With Shaft (aWork ) (b) (b) losses are negligibly Determine the head small. If all lossesP5. to 53. where 10 ft V is the average 3 ft 5.42 0 Q0is in 8.26 m!s.121 The turbine shown in Fig. 0.38 50were15.58 1a2 W1are!negligible.123 Chapter 6 Section (2) 1b2 2. T that12 the 0 0.72 non-uniform # 14!32 uniform 6. 4-in. loss determine from (2) to the the power bottom that of the the pump aerator   ■ Figure P5.56 Fig.120 of water is 20 ft /s.24 vu #is 90 !4ru mm.deliver de.94 ft the prob- 3 ft V 5.36 f # A ln r " Br cos u " C.62 5.78 The kW loss of available 8-in. 5.72 in ft when Q is in 1a2 3000 kg!s.10 in.114 is to are negligible. inside- energy associated with 2. Determine Water is the flowrate. 78. 26 m!s.148 3.at{K} tank ! {L/T a P3.8 lb. pe-/r ! r%2ing fcnp a2 6280 kg!s05.7 negli- are place pumping stations every 10 mi along 6. private es de {K} ! {L/T arrib Air supply: 3. umax /r ! z/Rr term %2 fcn initial mass flow rate into the del tanque a la boquilla está una bomba. 3.110 la línea que 3 the conduce x P3. (b) " " f 1y2250 during r/min day is at night.013 kg/s.0189 Pump performance 5.133an agua.01 ft !s.14 deseando /( ! fcn(r)comprar only a2 0.2) V is the average speed through the pipe. 3.66 slugs!s 5.124 0.5 km power purchase irreversible de carga produced del sistema 4.61 los 0. determine the minimum bine– om. determine el 6. It is planned to 25 $F!s 5.p/L)b 4 ] ! const precio alto.30 6.21 po- m real  en  Europa.32 f1 ! C12r.18 T$ Utilities! mean $ ! areL (∫uT /(L also dy)/(∫u %will-Vt) dy) .135 The pump-turbine system in Fig.1 $C!s 5.343 cosN(/r 2.875 & lb!slug.142 (a) que en la noche y. 0.6 N 17 ft.156 (a)turbogene- 102 kPa.3 5-80 La demanda de energía eléctrica suele mucho gener Pumpturbine H Chapter alta durante el día 3. atm. encourage bomba- P3. 2Vj]tencia .9°.126 1.often 0. P3.1393 1 # cot m.72 (a) 8. 1110 ft &height lb!s of 27 m from the ground. (b) at acros generadoras2 t!1 the ir rtance (or 3.140serQ 3.052 m3!s line leading from the tank to the nozzle is a pump.S. " 5–80 $éste !Q/2k rate hacia bh(1 $The Q %mese demand 3 cos /h (depósito will result? ) electric for enpower la 4. un empresario 0 Determine the 4.88 Stanton /z ! U ln number ! h/( (r/b)/[ln $Vcp) (a/b)] la energía producida durante el día por empresas privadas a un 3 5.  lion barrels per day (bbl/day)   (1 bbl ! 42 U. The temperature graduated 4.124 76.137.22 v # T !kW ṁR" 2 0' f 1x2 4.5 rad!s 27 m por arriba del suelo. 4 m .5 lb 6.143. laminar "válvula en el !tubo(32 4. En 3. If the water jet rises to a day.6 ftft h !consumi- 1.ing la pérdida 2) to .133 h !disponible m f # 21x " y2 " C.20 P/($'3D5) ! fcn[Q/('D3).43 ft & lb!lbm. System friction tencia y evitar la construcción de nuevas 3. where is sión.36 2 C ! %K $g sin%1(/(2-) 2 up. vencer la level height caída 4.82 in the kg/m 3 5-82 a2 15. 5.154 (a) 3.104 L '1!mil- ! %h (%V 1 (cot/R) ( )/2(1 ln % ṁt/M y ) del turbogenerador company sea is selling 75 40 4. Assum- cual aumenta 2ṁR la2' pre-3 # h)durantesellpipe el power día. 1c2 0 4. A.08 system rate' of is h 0.5 power turbogenerador ft. mand volume Respuestas: 0.25 1150 gal/min.88 Lake /z ! U ln (r/b)/[ln (a/b)] 15 m 3. ).andproduciendo discharges at night at itmuch Dpotencia through = 6 in lower acuando prices la nozzle. s 317 r/min avoid building new expensive power4. 3. P3. el 160 chorro rad!s.21 kW (work done water on is the also fluid) measured.118 3. normal. 1b2 53.042 m3!s A.116 3. de ! 70(b) tpre. P3. (V (b)f t/2L).114 5.8 3. consists produced during the day.158 (a) kPa. by theIntur. Si elestimate chorro de the agua initial A serate of hasta eleva temperature una altura de motor sumersopere The total to use como head theloss un available is 6. 674 lb sión 3.50 válvula2 ! Inviscid yUy K ln r la#potencia la ca 5–83 580 lb voirs to restore the situation. (b) 55. a long vertical capillary tube.72esLV)/( " de(a) #gd 8.8m.90 39.3 haft rota.160 .5 psi85.! 166 ft$3/min. (b) 55 m the aver- 290 lb!ft.106 H flow m de around a 180° frompre.100 No (a) $QuV 507 m/s n(cot and.122 1a2 4.     the flowing condition?   *P3.114 (a)2x414 3 r/min.p ! 0.76 de m 0.110 (a) 0. 4. For a design flow rate of útil e 5.106 'final2 ! 75 rad/s truir un depósito grande 40 m arriba del4. and 3.78 (a) V ! m/L " . 7440 m!s 1c2 116 N & m.03 dólar/kWh.108 (a) V !3 V0 /(1 # CV0t/M). estáp abierta a la When atmósfera.4 V 4.m.90 diam 3 4(H # " Water tricaat de 20˚C esta turbina.48 4.4 kPa. 92.140 3. horsepower needed to drive the pump.8m.25 3 de su 3.114 aumento h !r (a)!414 00vatr/min.78 kW Water at 20˚C trica de esta turbina. none e function 0 $0. z0 ! /H % *theR2s/r/ f # r " m Br cos u " C.rate 40 Q W is measured 40 m with 4. 1c2 22. the flow4. (b) p water livered by the pump? 3.5 3.de agua se eleva hasta una altura de moto 2 2 860 m!s .143 The insulated tank in Fig.33 trajectory of frictionless particles.34 N P3. 4.12utilitypIf! companies (p(0) /0%! 4.29 Selected m!s.18 Q .!V0# 0.sin (/r com . in P m.0204 lbf/in FIGURE 2 P5–80 4. P3. 1b2 146 5-79 lb Una turbinaDehidráulica = 5 cm Z = 150 ft 6.disponible show 1 in. (c) 710 !A ln hp r irreversible " C de carga de 4.28 porm 2 !as electric Exact %4Q cada a power ( solution/(+b)for una.68 are 2negligi- ! m tan [2xy/(x % y # a2theWlower 1 ! 9.03 ft (subcritical) 2 169.116 Will work for 2 atm.135 draws 3.21 (a) Vm wall.123.03/kWh of at night and is willing to pay $0.116 #P !4ru ! $Qr2! *[r9r* % cosQ ucot "(f21r2 constante de 3 cm.136 precio alto. Vf ! (2gh)1/2 5.110 734 ft city. 2 T each pump. 1b2 3. 3 ! 2am ion losses !185 kN.80 (a)ofwthe ! ($g/2-)(24x % drain x22) 2 3. ∆P = 2 kPa 5.837.  Munson. (b) cia the cc 0.140 approximated 1a2 cby upper reser- 1640 # the hp (a) 22 ft/s.66 507ft!s m/s2 and 1393 m. gal).135 riodos pico. is 4 14 m!s 5. y 0 0 0 5-83 a2 6. done bomba on the y de fluid) C ! $bh(1 anturbina. con- 3 .90 u # (a) dV/dt ! 0. 27 m .144 water vuoriented jet 3. P3. determine el aumento mínimo de pre.144 creates a 20°C6. which the w unction of 3.68 2 ! m tan%1[2xy/(x2 %5–84 y # 4.141. H/L) . potencial am thermometer. des absolute.!s 5. a y # 2c 2 y 1x 2 " y 2 2 5.135 draws water from 5-81 32 ft3!s 5. se experi.74 2 ! Uy # K ln r 4.36 aftprecios& lb!lbmmás bajos con el fin de alentar a los consumi.135 Pump ion Cengel factor 3.54 10 4.138power -for ! +$ a gd24(H2 # " Lthrough )/(128L presión "Q)aalong% .12 toutility measure m!s.66 % 10!4 m3!sdirecto hacia arriba. ULb const modos% ( $ K 2 de /2)(x 2 la pé # y2 friction head loss is 13 ft/1000 ft of pipe.88 1a2 231 N & m. Q ! 3 con un gasto de 0.118 0.86 4Q/x!!0.!for i 3 gasto de agua de 2 m /s en cualquiera de las dos direcciones. kg/s.92 dV/dt ! gh/(L # h)riodos pico.113 2 ft ) lbf. effici t with the kg!s3 !0.866 5. 10 ft!s2.   (b) 0.94 6 h ! 0 at t ! 70 s la energía producida durante FIGUREelP5–78 díaDpor = 2 empresas in 4.57anin. dólar/kWh. 3 ! 2am diam rador es de 78 por ciento. efficient and is driven CAPÍTULO by a 500-W 5 motor.4 ft!s2 5.136 m3/heléc- 5. $ Suponga que una compañía generadora 3.96 agua.07 m3/s 4. The water jet mayfrom be 6. " !y $evitar Q/2k la construcción de nuevas 4.52 2 ! the %4Q(/(+b) .  White.106 5660 ft & lb!slug P3. yAdejar 6 ft fireboat theque draws elthan agua seawater it fluya (SG del depósito 1.86 Q ! 0. y desea 4. The energía pump ais 0.20 m of the pond. What power must be 3.00 m !s 5.62 /2 ! Vy2/(2h) 2 # const of w on-uniform # 14!32 uniform 6.3 lb bine and (b) delivered by the pump. higher (b) at tpero apun !1 cieT del ! 20°Caguaand ! 200 kPa.ln(b) {MLT % EES ! 0. above the pump. f Para aprovechar esta oportunidad.30 h Q enp in!los ! Fig. N Si & m.6! 4y ft " C. and where its c ! overall la ftpotencia producida 0. *P3.kPa.106 gal).52 4. 558 generadoras Problems N & m!s pump Head. diameter Flow rate. las com- n.6 N & m!kg. (b) cm (b) 2.58 direc culate the rador es de 78 por ciento. 3 tanh Vpúblico /s.00456 slug!s increasing.20 m!s2. S 3.128 (a) 1.ex.88mar(a) y are V !la V "superfi- !#V[0"/(13##2V CV 1/2 de agua j] 0t/M).166 venden 26 kW Z (a) en la(b) 1  atm = 2560ft mi/h. The 2 2 2 moto !225 ft!s .0204 gz ! constant lbf/in2 5. If the generation en el pump ( capacity and to desarrollo del12isflujo 52 lb 27rise m porof thearriba air indelthesuelo.146 a 2los 3. 200 ft & lb!slug the upper reservoir in the daytime to produce power for a cons pres b2 2 V0 hb!3 5. x!L . length necesaria para and water L.10 3.142 rate (a) of 1150 2 gal/min. 6.164 frecuencia.56.38Lago cuyo Cz ! diámetro 1yx % 1xy2 es 5-84 surfa P3.2 3. 3. con frecuencia.7 lb 5-80 La demanda deturbine Pump- energía eléctrica suele ser mucho más syste .76 resultante (a) 0.88148 ft V! " # [" #0. 1b2 11. (b) 88 mi/h 4500 hp 4. (b) 250 r/min reservoir 40 above the lake turbina 4.92 T ! dV/dt ! 0 gh/(L 4. ft pañías de servicio público venden 227 m la energía Z 2 = 25 enftla 50 liters!s got to in. bine and (b) delivered by the pump.92 !36. 4.4° (r/a) % ($ his pump 3.403.0196 2 1r.102 1a2 at2.126 %15. 1b2 9.122 L! de-%h1 (cot ()/2 4.136 A pump is to deliver water at 20°C from a 3. The water density at4.60 this Irrotational. ! +$gd 2 m (a) 31 m /s.152 60.113 x!2 gft ) lbf.133 pe- ft/s.94 ft5-79 pressure Una turbina hidráulica 1rise supplied by the tiene 1 = 150 Zpump 85toft mthedewater carga line.137 barata.60 día Irrotational.0196 V /L. en () Bomba-is considering entrepreneur un 4. At night.174 supply durante h !5. r # A!B Pump– 4.0031fcn ($Ux/ 3 m /(s-) ) m) P3.383. and 2.133 .qxd 2/22/06 16:29 AM Page 223 P3.90 dV/dt ! g costosas que sólo se utilizarán un tiempo 4. 1b2noche 3. 6dethat the laminar 4. 0 Chapter 3.62 Because bomba-turbina 2 ! Vy /(2h) puede # const 2A Tom ho 3. System friction 3.138Students 1&(b) 1 110#4ft/s. 3.363.22 Inviscid use flow around a 180° 3 N Valve 6.100 a #(a) 5.25 ! cQ m /s. the friction head loss is sión FIGURA P5-78 5 5. . Estas compañías también están deseando comprar FIGURA P5-81 5. /(2+r2b2)] Reservoir a razón 20 L/s.112 2.128 the Second pipe. for gasto x # 2c 2 x 1x   2 " y 2 2. Se mide que la 4. z) Answe b2 2. to 4. and (b) the increases the pressure of water.4 $C!s.2 0.18 $ ! $0L0/(L0 % Vt) 260 min Suponga evated que unaThe compañía pump is 1generadora 120de ft/sservicio público kW is de.28 electric considera cons.16 3.178 3.34 p ! p(0) % 4-umaxz/R level pumping water from the mate the horsepower which must be delivered 3.0 deg.9 67 hp P3.10 4.89) T1 = 20°C 3.170 hf!un- !1.000 gal/min in either direction.128 1a2 0. con un gasto 1 D =de cm m /s y 1la eficiencia total 10 0.qxd place pumping mate the 10/29/04stations horsepower 2:25 PM which every Page must 10 mi 223along the be delivered  topipe.the valve is opened.204 flowm. Determine la salida de potencia eléc. level. !8.025) from de a!submerged /regreso al lago !and 2 5. the 6.000 gal/min in FIGURA either P5-78 direction.136 1a2 3.5 hfA!ln0. en la noche. km hydraulic turbine m above has 85 4.28 % 10!4 slug!ft0 3 & s 1 Constant- 2 3 2 24 2 2   P3. " C .130to the34 !yby oil the tube has generar porsuch año.pipe través 2$Q/(16 de isunagiven +L ) by hf.146 cby# h2x the ! 3. If the pump effi- is lo.74 la 4. as to! unidad in Fig.1.130 34 kW 0.900 and the lb %15.result: tempera.14 4. and Estimate the water thesurface power is in open kW (a) to theextracted atmosphere.172losses are plantasD 3. 0.158 (a) h 2.4 ft & lb!s Compilado  ejercicios  de  Energía:  Cenguel.120 (a) h2/h22 !ft/s.2 soor y2b power solution output for of any a a P3.82 water kg/m bomba. (b) 54 kW "Q) % .102 6. f stopwatch # !4x and ! 2y " C cylinder. Determine la salida de potencia 3.74 ft (supercritical) 4.303.                                                  Agua                                                             5.160 (a) de servicio fm 54 kW L evated vende tank.148 3. ) que ing P3.6 in the tank3xy k̂. 4. /[(.145 diverts a que usen la capacidad disponible de the river flow generación 3.p 2 /2 ! # 0.   son  iguales  y  representan   5.02 ft!s sión 1b2del200 agua.opportunity. f2a!large building z 0 ! el C2/r H % * R dete 5–82 / ft3!s 8.4 gal!min 2!1y pagarz2î flow 1 0.84 (a) /z !{ML %2 %2 ($gb2T/2-)}.0125 ort a per.126 No Pis1 = 1500 planned final kPato kW (work P3. it pumps water from lower to upper reser- 3 ne la a2 3000 kg!s. !Cflow .20 the during 2 de/! ! /Kxy day tuberías from # const 0 ! const.120 lower to 3. 100 $C!s 5. what2 20 N high-pressure air supply. el sistema The opera viscometer. del tanque a la boquilla está una bomba. (b) 209 m /h ! or 0. pipe 3.54 Q ! ULb wall.63partft!s(b) 100 5. !25.154 3.143 27 m3. 1b2 20 ft!s2. 4.8 plantas(a) 0.8 f(a) 1! is usually usando Cmuch 0.6 169. % cos cada uno. For a design flow 3. Un percent.5 pumps m. (b) 55 m by a Pump.8 ft & lb!slug Ing.48   m!s 2 5.138 3.76 (a)50. inve a2 10 ft!s2. 4.  Se  recomienda  escoger  uno.69 lb What is the friction head loss between 1 and 2.962x.70 head 0 ! loss 3 cos ( /r . P3. 0 m!s Write is 75 percent efficient and is used for the system in Prob.50 State 4.4 4. la 3.1/(ghSe very xsimple amotor )]y # 1espera #Suppose device a1b2 0.99 ft!s 300 5.25 m /s y la eficiencia total de su turbogene- . P3.118 P ! !Ar $QuVn(cot . .42 Tmean2! (∫uT dy)/(∫u dy)sureccaída de 30 thetravel uppera reservoir inhorizontal the daytime to produce 3. No2 nivel p del !lago.08 5  Ambos  ejercicios   20 L/s 5. P 2! $Qr dr0Z/dt! r # 2 2gZ/L 2 ! Q 0 cot ( only 2 /(2 + a r b short Depósito )] 2 2 time during peak periods.135 The pump-turbine20 system m in Fig.7 lb 20 m 3.  Waldo  Lizcano.38 4.00themcal.34 4. 4. El tanque 2a nivel del 3. 5–81 20 m Problems 203 pipe n. To take advantage of this of a tank. der a dam as shown.5Atank.70 F 0! ! 3. 1. 4xt d Z/dt.02 lb develop? P3.138 la p P3.(b) (b)68 88kN/m mi/h 4.108 926 ft & lb!slug.48 hp 15.106 m from A.132 2. hasta yused z) g # 2 *[r * %P3. ft3/s 6. entrance losses 4.6 hp sión suministrado What is thepor la bomba friction head loss a la línea between de 1 agua. # 2gZ/L 4yt .82 Obsessive ( ! 'R Jet 3.22 'D/V ! fcn(N. (b) ble. 1b2 62.2$Q/(16+L L)/(128L ") 4.4° turbine 4.14 C/ ! $ g sin fcn(r) ( /(2 only . con 3.4)  SISTEMA  COMBINADO cen72367_ch05. (b) 209 m3/h Water level 4. laboratory at Penn 4.(b) 14.78 1. 1b2 392the 1c2 675condition? N.104 4. del ideal gas.determine a 2 2 s.124 41 tur- r/min 4. given Kxy Determi- # const cm 9d rise cent effi. 5. 5.36 4. útil (b) 67de hpbombeo FIGURA diameter P5-80 d. the Esti- oil by ingre a2 200 $C!s. !2y. de 4.6 gal!min.156 (a) 9.152 h ! generadoras (a) f 85.108 z# 3.m (b) further constriction reduces V2 5. !28.112 6.168 3.137 inverso. Estas compañías también están 4. the de viscosity combinadas ofciento $x!Lde!la1. 41 ' The r/min! 75 rad/s suponepower quetemperature.55 ! ($m/s. 2 10 .89) at 1 mil.134 4500 hp nar. losses are it 6.a small diameter. u # p. 185 rad!s. f2energía V! C2/r /L. const bombear% ($K /2)(x # y . 4. and / ! /available head the 0 ! const. 1b2 W2 " 14.08 turbine–generator h 2 /m 223 Exact 5 durante el día.182 3.32 noche.08/kWh for and 25°C barrels U.96 !3130 ft & lb!slug supo friction head loss is 13 ft/1000 ft of pipe. not for 173 ft.1where # cot .100 0.134 A 36-in-diameter pipeline carries oil (SG ! 3.(a) 5. sistema stopwatch. 1b2 ne (b) 110experiment.84 z ! ($gb /2-) ln (r/a) facto % ($ tiene 85 m 3.144 pañías generadoras de servicio público 2 3.138.162 en the Q tank kp/[(k la noche. 5–79 (b) 14.22 % 106 W voirs to restore the situation. 45.58 % 10 m !s   !3 3 gene each pump.134 A 36-in-diameter pipeline carries oil (SG ! 0. head At night.92 1a2 43 deg. durante shown 4.6 4.lbm !ft 5.138 11. Initialestá conditions 3.791.7961/2 dores a que usen la capacidad disponible de generación de po- of kinetic . análisiswhat preliminar horsepower muestra motor is que puede required to usarse drive un it? sión suministrado por la bomba a la línea de3.36 ft durin 3 200 alta durante el día que en la noche y. g/2-F)(2 x %Nx2) !41.80 una   (a)! cw 1.   P3.24 P3.24 1m!s. las com. 3120 lb!ft to maximum distance. 2 (c) in 710 parties the1/2hpfluid mechanics at a high price. #&2&[1 %&2!63.136 water from 5.94# mínimo 3. 1b2 Yes.0031 ! %K sin ( /r 3 m /(s ) m) 2 a2 181 lb.90 F ! 3.2 4.48 (a) In 2a2! kPa. for Se any aatmó or b 82 N downward Tank : lion = 200 L per day (bbl/day) (1 bbl ! 42 3.134 8081a2 Answers 17.0 m3!s for each line 5. 5-81r/min 317 Fluye agua por un tubo horizontal.145dores The large turbine in Fig. la cual aumenta la pre- A dura 5. costosas que where sólo se is the average V utilizarán velocitycorto un tiempo in the 3.82 F onfiguración   ! 450 Nresult: /( ! 'R2/ Obsessive pump per. the lake ater.132(b) ft 55.104 ' Nofrom a ! (%V e /R) ln (1 % ṁt/Mciency ) is 78 percent.22 The $ head ! loss 1.176 102m/s. A pump is to deliver2 water at 20°C from a pond to an el- 20°C wa. (b) pmin noche a precios más bajos con el fin de alentar 3.120 1610 ft & lb!slug.3 3.9°. in m. 21b21 v to % 1 1 && #&&[1 # 8V /(gh #Selected 2 '2 " 2 Problems free )] 1/2surface "1 15zdólar/kWh is 22ĵ ! 1y'por 20 3m 22k̂.7. Estimate (a) the flow rate.16It3.122 0.8 m!s 3.4 8081a2 Answers h /h ! 0.32 ft. 6.max ! m/L ".144la energía 3.96 lb 15.343. right to left 5 bom !28.S.1 ft!s .5V2/(2g).66 remains! lami.03 eléctrica 1 m above the pond.263.16 plants C/ zefficiency !!y12will that yx%%3xbe xy 75 # fcn(x. are y is known. 1) y%desea ] # V.2 deg.33 ft!s 5. tank.183.102 0.que las eficiencias 4.16 / ! yprivadas 2 % 3x2y a#un   fcn(x. city. Because the tube is so long.6 m /h 2 It will be shown in Chap. que este and alevel. 0.42 4.134 de 25 1a2carga m Yes.162 1640 ! 166 hp másft3/min.6y head loss 26 !kW 100 kPa.12 If /( ! /0 !los corto durante 0.08 hp. la c #pérdida ft/s.112 tank is 0.max min 6.8 kN The pump in Fig.66 2 4.711.142 vende energía eléctrica a 0.07Esti-one m /s graduated ingreso cylinder.144 friction 19. in gal/min.143 is to be filled this72 percent turbine.180 pond toV an!31el.flowing N 6.. a # 4î ! 4ĵ ft's .4 8V deg. Estimate the power in kW (a) extracted 6. Inc. 5-26C PROPRIETARY)(1.   reservoir. = m the ⎜ = +4.un aestá control to reductor. de un flui. and the actual electric generation de la bomba es de 82 por ciento. Discussion In an actual system. de (50 wind 1031 30 m) kW is ∆T its=ciento. ⎟ =voir 7. de5 3cmm/s en arío elrazón de quelafluye de descarga. turbine-generator is puede given lasuministrar toflow be ρ rate: W de and =difference 1. the aceleración pipping power Assumptions ⎜ 1. =of A3 3 D326=/.072 in m/s.0between0). tocarga 20 m = If hasta = P V 2 air ve v la salida de bomba 3since asíthecomo water atefecto point is losessentially factores motionless. out ofinthe )the mech. 0).(potenciavolume Then the enbetween powerla m flecha) these used para two todel overcome points and themech.a windthis turbine change is must proportional be (9.7% exit ofeldetermined.0) 1 .overall eficiencia atm sonmecánica Further. La velocidad del aire no debe sobrepasar 8 . is irreversible pump .2maneras mediante 1000 tanque The Note una kg/m effect elevation ythat bom. any el determine ) Wind = frictional diámetro 2121 determine kW a) losses varía la ≅ in 2120de eficiencia (a) the the 20 overall pipes hasta kW total and effi- 80de m.V está gz 1 ) ⎟ 3 kN/mρ+ W1000 2 pump = = m.7preparation. loss L deη15 ft3. correction pool the volume 20 atfactor? aa given razón between arriba de Iselevation. electricidad. queware fluid. diferencia that la una estática.   m⎜ ρ + αVdesde α 4turbina.   normal? assumptions P3. the is deter !"#!!"#$% 5-66C ¿Qué es carga útil de la bomba? ¿Cómo se pump.= (de −4. Determine ⎛ 1En kJ/kg cierto ⎞ lugar lafluid energía el ∆T = 0.se1m and⎜ gasto kWesta- arriba is 70% máximo efficient. energy whichconsumed is bomba. 2 ⎞ 5-76 highe locidad efficiency. Wind as it flows the pump 20through m must boost theturbine pump consists of the change in the flow the pressure of water by 196 kPa in ρ 1000 Substituting. tres = correction E eficiencia maneras requires and factors hasta is pump 70% an is+ negligibl to be efficient.Th depó 12 Kinetic Theenergy We overall |Δ also Eismech. .100 kWyou de are electricidad. motor pe andW 1 = cuyo0 electric and frictional =diámetro ( 0 . 440P esfree 1m de ⎜α kPa. porhasta la tes 5–62 Inestá a contra hydroelectric arriba. ft irreversible Si3electric la How /s)(32. A hydraulic Considerey la El eficiencia 1diámetro The un río turbine-generator elevation combinada quedel tubo fluye ofCompilado   es del the hacia is generating deturbogenerador. ⎜ m + E+mech 4 αlas 4 + gz 4 +WW ⎟ subi subir =lam12 diferencia −deethe presión =Pentre ( pela2 yad.   kPa in three If⋅ myou ⎟ are different a student using this nerar Manual. cinética.través 7 kW de esa bomba a razón 5-32C Noting that ¿Cómo E pump.672 Respues- 67.0) las toberas de la turbina a 700 kPa de presión absoluta. actual 5-20 electric Se(flow m eva mech. are Chapterde 470 significant? la kPabomba 5está ythe Mass. is Suponga zero que 3 both 1 and ⎛a lake 21are kJ/kg detoopen a storage elevación ⎞to theentre tank atmosphere at entrada la a specified (P1 = bre rate. unto estanque theuse fluid a) iscuya energías.2% lico determinada 4 por el flujo en canal abierto? ⎜ 2 ¿Cómo se deter- 3 1000 4 0. take the = VΔP (b) aWe take points Discussion and pump thus as thethe power control 3 unaand 4planta atSubstituting.86 for m/s change in the kinetic (b) We take water pointsflow and 32velocities the and pressure +4 at on the ⎜ the difference inlet 1 pressure kJ and A the ⎟ acrosschange exit D of / 4 across the pump pump. de manera 5 Mass.m/s © 2006 ) The McGraw-Hill Com 2 2 rmine the actual Exprese la ecuación tres maneras (1000 kg/m 3 Under what conditions¿Puede do use incompresible. fiere respecto a la línea de energía? V + Eof V ¿En qué condiciones 6. 70 losses )( 3 se kW) re. potencia ésta the out kg/ma turbine razón mecánica 3 750 . V4 5 70 tanto Estanque =bine ombinedρW V max change pump–= =tro potencia. the 3Further. la= accounted for =by gast gasto th olution  nerador Assumptions 5-19 agua. de diámetro del del tubo completo es de en cm lugar.the tank fluid water = Bernoulli? mengz 2to. ciento which ofcinética dethe de power. to 452 de kg/s 2 this ageneración storage ⎟ =change hasta 2 tank 020.the 76 hp ⎛ 32. the may 2kPa. total are de cam. flow at the turbine 5-29C Discussion inlet ⎛¿Qué Pis (b) 1 (Ves equal WeV ≅ laV1to take thepoints presión ≅ ⎞ dethe potential . ⎝ ρ gasto Limited distribution teachers si and la flow pérdida educators 2only ⎝ irreversible velocities to for on the2pressure course de carga preparation. /s V ⎞ system.7% or 77.1ciento incompressible. 2. en 452 tric any la su kg/s energía changes de significado )( 5Disregarding 0.03 kg Answer: 444 MW pump /s.teachers MAQUINAS   5–26C and =educators ELECTRICAS   thefor–Bernoulli course Motores   13. of .  De   5–25C la  turbomaquinaria   What is streamwise (Potencia  acceleration? en  el  How eje). con el70chorro por laciento deeficiencia que de eficiencia delsale ge. and hacia 1 Energyde The arriba.the u pressure generation pump difference m potential motor electric . mass water. boquilla use determine que a) tank está energías.68 3-kW m 6 convertssubmerged kJ/kg) Analysis= 13. bothg flow Se water⎝ Equations is points la observa 2 steady level pérdidag (V que 1 ≅ aVtank irreversible el 2chorro ≅ incompressible.corrección en incrementos 750 kW dethe the turbine Analysis  become (a) FIGURA We take P5-19 the free surface of the lake to be point 1 and Wemáximo the choose 5-67C free 70 the surfaces m points ¿Qué free es surface of and the factorof2 at storage the de the pool. gz/34⎟ += W339 3 /s . α irre. power can be=lagenerated eficienciaby mina amla= salida P P 3 = ⎜ ⎟ = 7. (V1 despreciables. sale elevation equation ground. ρWeun take Vflujo2=thedepump =W fluido ⎠pump. lake 2 Trace are and cinética? m open la the 3 pool. de flecha) ¿esflow pérdida de exacto10water incompressible kW.86 m/s = (0. 6 + 343 . take de de usteady W ofthe la the energy tuberías bomba.= ηson nivel The en piping ellas del pump-motor de tres agua la = theelectric m motor W hipótesis pump.70 Determine = side and 52 cm on the )( 3 está kW) power.07 m)(2 3. loss. 452 This Wproblem kg/s ⎜elect )( = η wind 0. − es We u la = = takeubicación m gh the .0 ground.7 mech = kW: ( incrementos bios analice 29 . Wind inlet by la and bomba. and thus 5-69 the potential 1 unit 5-77power mass. Chapter 5 Mass.672 la or densidad power. The mass flow rate: es from The como actual 1. tic energy and the iskW ne pe actual the lay=only eficiencia to lamechgz electric de generación =form agua (9. ⎠ 800 fluid considerar generación energía 41.7 we por considerciento V the pump.   ybomba-motor 5-63C una descarga bomba Considere(potencia dey la elbomba. y delendiáme. out )fluid =Estanque m(ispesupplied 1 − 0) = mpe1 hen the rate at which thefluid mechanical energy of the the We mechanical take point energy 1 the atloss? the free surface of underground tica. de P 2 la versible 5–64C The =laProperties fricción P de ). blowing steadily at a constant uniform velocity. fluid En una planta Equation mech. la admisión 1000 diameter pipingm 2del of/s 2 ⎞ 5–74 η ecánica = viento drives to con turbine masa y el potencial de generación de potencia de una turbina de the de álabes generator.un piping líquido? 2 turbineW max location 90 m menta = ( 0 . del mech alloss. rate turbine En of de increase cierto esta planta. 6Pump línea kJ/kg) take 2 de mech.steady = m is de gz of manguera agua . después determine Energygz ⎟Also. pump de +resul E me c ca delorge. where ow strongly the free sur.0) (1. puede to re-This lbm/ft reduces be neglig is theto) he density combined airque ofunit forme. = in 2 . Further.4 blades. fluid mecánica de almacenamiento dicha mech. fluid = effect8 P.2452 pe Respuesta: = (turbine. . head? of theat wind How both is turbine it locations related to is are negligible the Chapter (V We also d is blowingindependent   take steadily at of the FIGURA lake a certain surface the windvelocity.2 kPa ≅ 289 kPa l turbo.0291 and m.⎜14 × pipes 10 y reduces 1 2kJ carga m to ⎟ ⎜⎝ ρabout /s de turbine head bre hp. gz loss ⎟ + W and + E gráfi Wind m 6 mech. )(in12 =una ) m/s) m(Determine pe 0)de=el=mgas- 2 −energy 29to.W 1162 α2 4 +kg/s + 2 E+me eo free la del surface energy2 5-24 tro of cubierto only theSe since reservoir. 82 1V surface Bernoulli. se Bernoulli.lb th1o ak ore. u m de⎜ un ⎜ = ( − 4 α.beentrada point 1Properties misión and y thela versible descarga turbine de carga de exit la del to bomba. rates pumped density loss of by water of amech 70water como loss. correction 2incompressible. u be potential its the converted definition.93 +u 294 +pump.14 la +atmós-α≅4×289 + gz 4 ⎟ + W kPa ⎟ mine 470 turb (b)ta:Now 64.687 exit kJ/kg of the depump co. pump mech loss. generated of increase de 20fricción by m. toberas used of the to pump lower overcome reserv nera it Solution motor? A la hydraulic turbine-generator is agua generating electricity que se cenamientofrom consumenthechocar deque water 20. converted mEl h respecto pump. Bernoulli dey adiabático 36 ft. fricción 3 determine All en the los losses a) lain tubos yeficiencia cualesquiera the pump h pump.⎟V the pump 2⎜ V carga ρ de and + the ⎜5velocidad ρ 27. and eficiencia the kinetic mecánica Assumptions varía u de es 0la energy de latemperature 1bomba. tuberías entre las superficies 2 libres de la fuente y el sumidero es 5-41 turbine with 50- ubos. the ower generation kW of actualhp. 2IfWThe 5-72 Wth 5-18C unDefina estanque la eficiencia cuya superficie de la turbina. =5-70 Vuelva a considerar (1000 ⎜ kg/m elVproblema3= 3 )(1. los álabes a razón power de laplant.589 m/s McGraw-Hill = Companies. pump is negligible case bina (or EES de other) 5-64C measured.)(30 m) ⎝ ⎛ 1 ⎟ ⎠k 12 m/s menta 0. con la bre la eficiencia entrada Analysis demecánica potencia(a) a The de lalabomba?bomba.5-25C per ¿QuéE Underground mechmass. 30 ≅ taking 2120 )( 2121 point kW kW) =1 to 636 be kW at the   flujo 5-29C turbine inlet. de de dicha co. manguera are 2 V at to dethe la2 be 15 presiónfree al m. it these two Th kineticción Properties energy de un at both turbogenerador The points density is zero hidráulico.and Respuesta: and The 78. © 2006 The McGraw-Hill Companies. 31 3en− We35la está ¿Puede línea pump assume = 20 29. volume. inlet constante losses u and the in parapiping outlet un and are the disregarde 78. E gzvelocidad + W = es =EV m y ⎜ 2 elevación + α + es + z. Equations tubos ySe cualesquiera descartan cuales-cam. m 3 /s 4 5–76 h = me 13. eficiencia = which total generates de 30 4 por = 750 ciento. ⎜ ⎟ = 196 ⎟ Respues-kPa mina a la salida W pump. para (0. of W ⎜air y5-70 the given 2to⎟We 0kW .116 × 10 m /s − las 3 dos 3 máx ge.  Waldo   misión unidad Lizcano.072°F kinetic mechanical Tome energy. z un presión ) 2 = normal? 1   water loss.ΔE Simech. la 2 generation diferencia )( fluid 20(12ym) m/sla energy ⎜of the ) eficiencia de are ⎛ to presión 1 the bekJ/kg determined.115 . flujo estacionario the kinetic We uny choose energy adiabático points de un and un2 at flui.1 3 100 de+3kg 8 promedio 5-69 la de Seym/s superficie 3 b) alarazón quiere diferencia del bombear lago. The 1 po) subir mechanicale aguainvestigue hasta energy Wel una efecto (62. to the useful mechanical energy en lossupplied yby the pump. pipes. the pump1 kJ/s de mass la as línea and the de volume control m gradiente flow volume. promedio ydel detanque.motor 50 mla bomba.7 por ciento fluid Welect. u = 1000 la pump both points unit and is zero pe the1ypressure = (ke gz 11 = Río = ke (9. de Respuesta: la admisión 5-75 Vuelva aatm considerar el problema Tanque 5-74.Also.ρ tank ×⎟⎠ 10 mp the motor 1 pump 2 2 turbine mech. 1 of piping kW the ΔEpe2 = fluid gz2.of el out = = P5-69 0. (P w . =1 0. pérdidas consumen por 20. (1000carte and generation ΔE mech. = (salida 1000Water kg/m is the pumped 3 )( power 0. ρSi V Therefore. of aelectric otro Determine power. m/s diferencial and (b) en Pla pressure )−admisión mdiffer. se piping apunta z directo por Analysis determined ción abajo de unfrom de turbogenerador la their superficie Kinetic definitions. the loss L flow and 2 3suma Bomba 2 herefore. ⎟ converted pump to energywork 2 ⎜ ρpumped 2 2a for ⎟ ind speed. ⎠from del 1reductor.4 pipes se1. determine la velocidad máxima teachers and educators 3 for course preparation.   to at desde atmospheric   de   the un lago turbine presión entre pressure exitgrande is of ladead- u hastakPa.5 incompressible. = m. density adiabatic pump-motor u Suponga es of de flow water 7 m. de is theenonly depósito un sitio formque of está de mechanical agua 70 m Assumptions Ecuación energy the de wind Bernoulli possesses. u =expréselas control para volume. of the pool. energía pipes is 3 accurate and   Theare el effect problema mecánica? to negligible. and outlet agua m Se 3 is/s.9 + 294. Turbine es la loss. − la = m5-31C 0) Thus. realis13 de=zero . nivel deV 2 theFor flo Wind is blowing kW forme.conditions. 70 bothm inlet subir 0. the electric energy P3 = 5-73 by teachers P4 −consumed the andseleccionar educators for Se pump-motor debe is uncourse preparation. kg/m 3In (b) We take points 22   2 3 teachers and 4 at the and PROPRIETARY inlet and the exit educators forofcourse MATE thelos pum f energy only since the elevation difference across the pump and the 5–70 Discussion 2 . are de the la aceleración 2006 three 5-­‐69. V suel 3suelo mechanical 13. 2121 072 solved =kJ/kg kW ( 0 by . 100 mung erate ρ negligible.   © 5-­‐21.68 elof determined. specified 5-705-25C is negligible. =Para MATERIAL. and el 2lado are open to the atmosphere 5-64C (P SeConsidere = Pel ).W mass. Considere el flujo estacionario y adiabático de un flui. la W flujos 5-31C pump.81 − 2se⎞ deter. /s  to Takethe re a Assumptions 1= (The 1500 wind kg/s)(0. de Inc.  differ 3 pump erinstala- output of the velocidad baja. 15 Theof water flow m. 8 kPa ≅ 340 kPa m ! 3 m ! 3 m.141kg/s kJ ⎛ P ⎠ V1042−3 m 3 /s ⎝⎞ and 4 ta: 64. of 0. la depósito energía quecinética.64 V m/s P4 − P3 =⎜ ρ © 2006 ⎟ .81 m/s )( 70 m) ⎜ ⎟ = 0 kJ/kg Substituting. W ⎟ + gzloss3 pump. Now  5-22 2 Itm would we =⎝ 1000consider ρVVuelva ware =give m(1000 2 de2pump. boost estacionario? the If youpressure are a of student water by using 196 this kPaManual.2 ⎞V 2de lindde pump. pump. ih rbine exit is   assumed to be negligible.−potencia kg/m e mech. es energy determi- Determine de 800 Solution of gas. resultados Vuelva is flow.072 remain kJ/kg constant. uno .07 uno EVmech.generadora out mech.2 constante elevación manguera ft/s para entre se ) la (29. partícula 5-69 and across ofExprese m 1 Thus. = loss. 2 Determine a) el gasto máximo do pressible W determined. power a energy generar in e mech. Si las salidas de las toberas están expuestas a la PROPRIETARY =2 =eléctrica. =gz1an Then ⎟ +additional W pump m = +m 2 ⎜ energy useful⎜ equation α 1 ⎜+pumping +αgz21 ⎟ +−W ⎟ for 2 steady 3+pump power ⎟ gz3 ⎟2==⎟m+6of incompressible ⎜ . which boquilla bomba. fondo⎞ Si surface en mass lala del eficiencia of línea flow tanqueundergroundcentral ⎛ y P mecánica 2la boqui. de turbina. P5-19 as the reference speed.072 can kJ/kg also ⎟ = ) be W =0 .64V ⎛ ⎞ 5-402 The power   generation 5-25C ¿Qué es V ways of a (a) wind energies. turbine becomes y ρ W pump 116 E mech mloss. volume between incth these (b) does2 alamechan. está u Can de 25E Si pump EES mech laW m arriba. are to be determined. 2 the flow rate of water ⎝ 1000 be kg ⋅ less m/s because ⎠ 6. piping equalpotencia. pipe energy pump 7iscm-de isconecta to aton work points 1Wpresión intake and 2 Bernoulli? are pe1)(30 = 0del and ⎜tanque discharge ⎟ una 2 pe también. carga Consider del W efficiencyWekinetic despreciables? the sistema the pump. surface and (PEecuación gz using 2manguera ofde Pflow = maximum de (9. el consumo de 13. bomba indican que la temperatura del agua au. unit hidrostática.14 ρ kg/s FIGURA FIGURE 2 P5-69 P5–69 ⎠ PROPRIETARY ⎝ ρ   2MATERIAL. Bernoulli. We the andalso kinetic Energy take = energy the = Equationsof water = at asB)   theFreference lujos   clevel on  (zEto2diferencia ficiencias   epotential n  potential y de 5 cm en el de 1 la descarga. Se eficienciade observa 3 kWde yque la turbina. 3 de We assume the frictional 1effects pipi rrección El de la energía ríozero cinética since son both ese1en 7 despreciables. bydynamic. αde 2hp ntial. mThe 30.035 locations suelo. lado point mech The de⎛loss.and Si el consumo de potencia por Storage Tanque tank la bomba es de 27 lico determinada W por (1000 el flujo2kg/m en 3 3 )(1. of Also.68 is blowing 7 kJ/kg) steadily at a constant uniform (P 5–66C velocity.25 también.y Determine con remains 70 porlaciento constant. decir irreversible is Sipumpedlaque delos pérdida flow carga to athrough efectos irre. kg/m energy. both point presión (01potential = 1kg/s at and . u ⎝in cause antes the2 pipes yandthedespués aretonegligible. pump del cuyo /flujo? 2are Discussion Explique andfor and intocomo diámetro the piping the be ¿En qué =nozzle a determined. máximo 5–68C mente.becomes a pump Analysis 5-26C losis negligible.1pump 3estancamiento? and 2the kJ/sP41 atmust ⎛energy ⎛the at PV raise 2inlet 21 ⎛ Explique 1000 and⎞the V m 2 2the water /scómo2exit⎞ ⎞an of ⎛P additional the pump. pump-motor Suponga draws que ρgzpérdida 3-kW la 2 of power. / pump 6.12 43m) ) the +kN/m 2× 10 motor. 14flow × rate 10 −3 rate of m 3 water /s of will Resp Resp 5–78 b order to raise its elevation byBomba 20 m. kJ/s oequation  Generadores-­‐. 196a del otro turbo- kJ/kg lado and de it can la qué be 0.level. = 196 kPa ⎛ P (1000 V 2 kg/m3 ⎞ )(1. steadily Se   strongly at recomienda:   qué with difiere 5–27C MATERIAL. L udraws en un = ( importantes 0 tanque 3-kW .14 kg/s ⎜ If the =this wind deturbine at the or stated conditions.volume con lalaubicación =flow V usuperficie mgzrates 3 = 2 de =libre water =ubecome =⎛ 11000 . turbina kW. 4 ⎛hose wh 12 Discussion m/s Theenergy powerare generation negligible. 70 Bernoulli por 3-kW energy Epipes. . libre está la eficiencia30 m arriba del ge.   does  c)  Eit. 3 + α 3 and + thus ⎟ + the power = ⎜ 4 +   α 4 + ⎟ + 1 kN + de 800 ⎝ ⎠ the P m − 4 ⎜ 3 P = 5-72E gz Vuelva PROPRIETARY W a considerar Discussion m MATERIAL ⎜ ρ el problema In an .1 kJ/s hidráu- 2 ⎜ 5–77 5-77 ⎟= =D3 / 4(1.25 elect.6% diferentes. the debottom arriba 2 la includes frictional la outlet of dirección horizontal. the respecto absence of⎜a alaturbine.E additiona Properties ΔWe E |mech. be sistema de 2. = nivel Si to h u de se del work la mide pump − agua ρ ( aceleración pump zque is − en la mech 1 negligible. por + 1α ciento. 0. and deEnergy carga de it agua. mechanical kg/senergy de Use el Soft. incompresible. the Se Vthe Se The pump wind conecta /la quiere 2elflow W flujo ecuación for is isdensity bombear turbine una adetermined. cm dethe agua lahose mhasta de ⎛ 0Pla =water aceleración 4pointed manguera 8Generator 2 straight V1normal? se to ⎞¿Puede eleve up. que of to labepérdida determine an 1 kg/L pump. presión atmosférica Solution Water de 100 isarriba. la aceleración turbine is 0. = laBomba EESWe (o decualquier7 hp take (potencia otro programa density en water of la flecha) Wde to este ⎜head be upara 1 lossti- kg/L 1000⎟ entire 1 = (9.81energy =ism/s) las que636 pumped m/s álabes arekW )(from velocidad 20sobre negligible. las energy. a in⎞considerar 20.es water.115 Determine kg/s 2theteachers m/s 2 and m/sis=) 1negligi el2educators gas. The la fluid = generación E flow mech de 50 mech This la = energy pérdida problem me real m atpor de = can both ( irreversible de 29 diámetro potencia also . b) la exposedof water Mdiferencia agua unson. 9 + 294 Noting that . y water at a rate 1 3 3 4)(1. incom- hasta loss mech m los ! neración agua. 1 and   ispe steady 2 = 0. surface motor andofW its the FIGURA potential reservoir. Bomba in the a considerar is the Bernoulli. The potential ⎝ energy of water at point 1 is 5-65C Analysis¿Qué m (a) pérdida g The irreversible ρ V g de drawsSubstituting. 2Note Also. an 1000 actual m 5–69. uni. = m(e mech. the Desprecie viento mechanical está las energy pérdidas soplando of en water uniformementelos tubos. ⎟⎜ W The Determine the McGraw-Hill flow E el rate of +water 2 ⎟loss. energía 2 The mechanical de mecánica eficiencia hasta energy el curso ofEquation water atExplíquelo. ⎝ V3 = (9.flujo? ¿Qué effect flow and esusing thethrough la presión kinetic flow gz aenergy de energy+ gh 2control incompressible estancamiento?volume L instead gflow ( z 2 of +through between hfactors) a to LExplique these control becómo two negligible. de 500 agua de mpresión Determine del 3/s en laun subsuelo deenergía unomediante lugar a90 otro mecánica m lado una bom- por de la do accelerated the incompresible. 2 gráfica 5-75stant laciona Frictional Chapter it 20de con m during delivers 5 los pump la losses Mass.5 V 2 (12como m/s ) 21. fluido In⎞⎟the A3 absence ycarga D32 ofde expréselas / 4⎛aStorage P velocidad turbine. cualquier de pipping ( 0 aalmacenamiento. loss que the piping puede FIGURA system acelerar and P5-73el the mechanical medio surface power las pump. 1000 Discussion equal generation to the useful Note will change that only mechanical strongly two-thirds energy withsupplied theofwind the the pump.05.across 7surface by 14 the the efficiency.= así free como el Wof efecto the = pool.05.0) canal abierto? gz⎛21000 ¿Cómo (9. extracción deWde tanto alma. tolacan a través ⎛ aP1pool de atVa1cent. !!"#$%&$' potential.8 kPa ⎜ 4 + 4 340α 4 kPa V4 A4 +   gz 4D boqu ⎟ 4+ W 222   ρ 20 1000 m kg/m ρ α 3 2 ⎟ pump = m⎜ ρ 2 ⎟ / ECUACIÓN DE CONSERVACIÓN DE MASA P2 −ofP1the windWe ⎝ Pump volume. (11031 .0) ⎟==(165-69. In ispor an arriba acelerar-water aactual ⎛ razón Pdel system. In endthethe absence head of aisis loss turbine. +udel W E mech Pel 2 gasto both reductor. 2 l D efficiency ofpo)e a investigue cenamiento que 3 elestá efecto 20lade m 2(50 la arriba. u = cuyaW presión − pump⎛ P mech loss. 3-kW ρ se 1power. 70water L/s. 2 g correction = at2!"#$%& which    the  ⎛ mechanical energy ofm the fluid is   supplied ! !"#!!"#$% used 𝜂!"#$% 1   kJ/kg  𝜂⎞ !"#$%&' 90= 8. la en u pérdida b) el extremo presiones 2irreversible de y c) dichacargas. la turbina arede manera es de 800 uni. the re.1to =Assume ofthe presión kW underground atmosphere water.25 . 3 Analysis total del bomba.070 frompotential 3 =lake ofkg/s 70 to the a storage istank its kinetic at pump a Ecuación energy.14the is× 10 pump converted 3 as3the de mcarga /stank delthe sistema de tube. ρdetermine del= 62.de carga manguera en éste.     ⎛ P3 5-71IV32 Envelocities ⎞ V across ⎞ the ( ) ( ) 2 2 ⎜ ⎟ on the pressure ⎜ change ⎟ pump − is 5-56 negligible ⎛ pui Eléctrica.4 and fricción kW Energy de en los electricidad. byVsi the la pump-motor V Weirreversible pérdida take 7. kg/m . 30kg/L m arriba de dicha 3 ΔρEreservoir 5-27C Properties ¿Cuáles We take son the las density tres hipótesis of water importantes to be 1 que = se1000 esta.motor head loss? is it related to the 𝜂!"#"$ = We take Properties       the density   of water  𝜂!"#"$%&'$ to bepower ρ== 1000   3 !!"# kg/m . become 1.laThe flujo. ⎜ ⎜ un + flui.64 / 4 m/s 2 )m/s ⎜ ) 1 kJ whos descarga de unasubstituting. by since del si se kJ/kg Determine takingboth ) supone = point 1aire de 2121 and We tube-it 12toare kWblecen delivers 5-28C take ≅ beopen en 2120 de point atmass. In the flujosabsence ⎜ m⎜ presión cuya of a turbine. (70 ¿Qué kg/s)(0.589 m/s MATERIAL) − ( 3.81 de power of y la m/s V turbina mechanical mech. V velocidad de 21000 pump . delivers Equations varía de 10el to hasta 15 fluid is m.776 72. instead motor a W rate the m) y otroand ⎝ presión which ofof2water electric lado = ( volume 01. a low control velocity. agua We Suponga del take río por the que unidad free la diferencia surface de masa ofde y the elevación el reservoir potencial entre detola ge. cm enis The mediante pumped la effect 1 The dirección ofun flow ⎟ athe reductor.19 = es m ρ(b) gz la2We . 86 m/s ⎛ P 2 − 3. agua = yde la 50 which diferencia m = de 0. 67. deefficiency 1kW de 500 kg/s. pool del kinetic at steady flujo? aand given energy ¿En ⎜ elevation. Ing.(62.070 change el since Pump m /s)the problema 3otro in programa the = 70 5-21.116 ⎟ asythe loss kg/s mech loss. and thus the potential energy Energy ateficiencia points 1del Assumptions and flujo? mecánica 2 areuseful de11la pe The gzflow =pumping bomba. V b)E. Tome por la to Chapter pooldifferstea at 5 ⎠a por 3 m =abajo de la superficie libre de mun /s)adepósito grande wind de agua 2 Solution entirely. the of thecarga? fluid con.P5–24 the remaining one-third is wasted becauseVof= therías m V 6 =. Then thelibre ⎠ de mass andun volume líquido? ⎝ flow rates of water become ⎠ water ecánica e at an average descarga de una bomba indican que la temperatura ρ del aguayau. We tuberías alsoes de take7 m. 82 turbin pump.072°F Δ conforme E fluye 13 a .4 of altura u air =η lbm/ftdeper dela315 pérdida unit )(1.W pump = 5-69 Se quiere del subsuelo mediante una bom. in you are using it without permission. = fluido ρ ⎝ lbm/ft en 2 .727 = or0.7 WkW pump 7(a) to a itThe poolpump-motor whose free surface draws 3-kW of power. el agua entra   R ecordar   l os   t ipos   d e   e nergía:   a )E.070The m /s change ⎝ 1inkJ the ⎠mechanical energy fera? m= of m 7 =⎝. pump. Then 5–69 factores the Then power the mass used Properties corrección tob) and water overcome volume de is la toit We energía flow be becomes take mech.We of hasta Si take essentially conecta the se ⎛ una pump point mide una P altura 1 que stationary.2 m u = gh energy Pdeatm al m/s η ). kJ is y de 70 )( la alsola 3 kW) flow ⎛ ⎠ taken = rates If 2 1 . that both 1   and 2 are open to the atmosphere (P 1 = P 2 mayortential energy. that Equations kinetic 3 the Use g energy el frictional Soft. =)(1.3 mcentral acelerar. 2 The efficiency of the lla wind que turbine en el is extremo z dicha h manguera loss. distributiondetermine el permitted gasto mínimo only to necesario para ge- the de agua combined LAS   Ecuación T URBOMÁQUINAS   a lade que Bernoulli ΔExpress E mech. = ( −3.137. energy libre hidráulico. a)que=20el 2 . pérdida to negligible.Dthe power 2 side. be5-70 Bernoulli. the energy equation ⎠ ⎝ ⎠ power by installing      a                of  Δ    E  a    wind generation      FIGURA            turbine        =    P5-24  m      (  e  is      proportional                  −      e       to)the =m cube velocity. and enunidad incrementos bomba-motor de pump–motor 20 ym. the kinetic de ti. Bernoulli. 6with the Bernoulli 5-62 wind mech. thegeneradora Noteand inlet that frictional losses el inagua the fluyepipes causes th ge ⎛ P4 −3 thehidroeléctrica. b) la flujo en la diferencia estacionario flecha) de   de presión 10 y kW. 73 razónotro (P 0 of )( .589 116 = ) − ( 3.25 the kg/m mecánica free out surface delof aire 5-69. de un tubo que2 está descargando a la ⎟atmós- determine mech. ⎠ men Cengel 05. combinada wind de = ⎟750 kW. both 2 energy Fluye input FIGURA points thecm atm agua to at P5-24 power thepoints (V en W pump? 1 un pump. kJ the 5-9del ⎟ cube flujo? of ¿En the wind ⎜ velocity.81 difference 2 =la0) m/s )( between70 m) ⎜ theelinlet Estanque andde ⎟ 1the= 0 .|fluid speed. respectively. con una V+ una gzaltura 3 ⎟V + Wde 240 6.072°F30 )( 2121 conformekW) = 636 fluye kW a través de esa bomba 5-32C ¿Cómo es A la línea pump. 3W tuboVque = ⎟descargando = = hp. condiciones dos piping and u pump 0ricmpower por generation 5-23I ciency Un ofΔthe sensor E mech. kJ/kg mmust en be 5-30C de elevaciónDefina ⎛para P1 un carga flujo V12 dedepresión. is theirdisminuir sum constant for a flowdel fluido en ⎛ 1 kN ⎞ 2 all menteefficiency diferentes.locations 5–71 turbine-generator nerador boquilla y la eficiencia se efficiencyeleva 25 combinada m and por the arriba del turbine del turbogenerador.7 gradiente kW and mechhidráulico? 4 at the ⎜ ρ¿En inlet and qué 4loss. inlet by Weand FIGURA theoutlet ¿Puede take nozzles P5-73 thepointof before the disminuir density pump striking oflawater are temperaturato the tobe be determined.qxd 2/22/06 Wind Pump 6:29 AM Page 222 V gz=2 + gh = L g ( z 2 + h L=) 7. 0).1isof be el2 + kJ/s level.07 P3 343 s ≅and V . Determine V the the flow 7 . and thus the power diferencia de presión 3 Storage ⎠ de uno a otro lado de la 3 5-78 50bymelectric versible head loss of the piping −system 5 m. e =dedensity laaρtake elde3problema The Therefore. pump Com 2 de agua⎟⎠permitted 3 3 pump 4 4 turbine mech change wing ROPRIETARY determi. + α 1 es +P. se puede – Turbinas   acelerar o   el agua B ombas-­‐   ⎛ 1por Y   D medio E   L AS   3 ⎞de las toberasWe an. de dinámica fondo pump.adiabáticoentre Siturbinela pérdidala de m ad.035 del 5-78 difiere tanque. 86 m/s ⎛ P≅. y kW que de 4 el electricidad. is sistema ⎠elevation del 2 /2 per 2turbo-The deunit tube-   difference efficiency 5-26C se mass.pump-motor en incrementos at steady Respuesta: is 78. stream given η turbine and themass pressure = at =between thees=de 0. +conditions.4 kW 5cualesquiera hasta of elec- 20 m/s. The determined.12 . and which frictional alsolosses is points  that5–75 taken in 5-75 as pi me mech = =scussion a-motor =( 29 . water E Bernoulli incompressible. − ( z 2 which −fan z1 )c in − e mech. 5-24 que = puede 1031 kW suministrar ésta FIGURA a razón P5-69 de 1 500 Chapter kg/s.freeenergy are kW) surface open 2y. Substituting. la12generación )( atΔ m/s) E It3mech. acceleration? DIFERENTES  DEBIDO  A  LA  EFICIENCIA  5-9 Can a fluid particle accelerate in DE   ⎝ potencia 3 A3 D ⎠ /42 una (0.141   ¿Puede used inacelerar- Companies. percent 1 kg/L mech los =1 2 (12 m/s 12 )5-21 2 m/s.70 the dinámica )( 23pump.14 × 10 m /s − 3 3 kinet m water Therefore. Underground⎠ is Then 20 de de m ρ gthe carga agua above water que 2 en ⎝ 2 The the is g éste.07 the if the m)pump irre-/ respectively.negligible.2 =V(power eficiencia the = 0ft/s.4elkW problema 5-21.efficiency suelo. the which + αstated 1 1 requires 2 0). los Vtubos. la admisión de  Ethe 2 The nergía:   unidad water Ecuación 5-74 mechanical nozzle C 5-69Seoutlets ofenguel.4 m/s.589mm/s /s .2 m) 2yEcarga de2estos /=4⎞de E mech loss. el20flujo de estacionario m. P5-23E energy at point 2 are both hidrostática. the fr 444 MW 5-745–63C Consider 1está bombeando the steady agua adiabatic desde almacenamiento flow lago grandeof mech.⎜2a3 kPa kg/m 289 ⎜ 4 7.   + gz2in2fa termined from their definitions. the turbine exit is 5–72 turbine exit yunlaagua as salida the reference level así (z 2 = 0).776 generates diámetro de or presión en 77.25 the waterkg/m 3 at . 3 uno s2≅ Noting de estos that −3 z = 3 z . Note that b) presiones only two-thirdsy c)la of temperatura cargas. point = η and 2. Explique are to Chapter qué be puede determined. msince pe1 de both 1 and 2Solution la instala. of air (ke is 1 en un = given ke = to 0) be since 2 sitio que está 70 m the ρ = 1. 2 . determine ΔP = la20eficiencia m =in mecánica 3 de ⎜ dicha bomba. the qué air difiere pump are duceatunit the Respuesta: causar que torespecto de be 15 0. order to 5–28C raise 5-63C itsDefine elevation Considere static. eléctrica mecánica el2 problema The de lamechanical 5-69.una Assumptions suelo. velociti poweener m/s rbine exit 5-24 turbine exit de agua is assumed y b) la = be0). the di- 2 exit of the ⎟ pump.729 potencia .6% 750 ese de kW uno lugar. volume the pump. Trace +ofα la1   +2zand el The vo 1pump. points= 13.. ≅ 1 Vand tubo u = 2 η ≅ 2   horizontal.6. Energy say of the5-74. 4 ) kN/m = 339 .81 ⎠ m/s = ⎝)(30 ⎝ m) = Pool ⎝ ⎠ ⎠ ⎠ = 1. gráfica ¿Es = P FIGURA de = significativo? P What los P5-24 The 2 )resultados and is useful efficiency the pump yvelocities analícelos. generation atpotencia outboth are meléctrica to be 030)a m determined.cubierto 25 )( the 15–24 kg/m 2121 =kinetic gzSuponga kW) 1Water = )(12 por (9. 452 points kg/s en be is de )( carga eléctrica ese solvedzero 0. the fr nalysis at 444 both WeMW take points the is free zero surface (ke = ke of the = 0) reservoir since the to be water pointat 1 point and 1 the is Chapter turbine essentially exit5 Mass. ⎜ 5-69 1 and ρSolution itincan 1 2and be pump. ρ thatcarga includes 2 de presión.2 lbm ⋅=ft/s del2 v⎞ pe1The = to gzchanges 5-21turbine the in !kinetic become energy are negligible. 5 quiera que Mass. | Δ E 2 | ⎛ 1031 1 kW kJ/kg ⎞ se pressures una from partícula at the W the 3/s. the to suelo. the /sthe turbi a kg/m EES ne ⎠same )(result considerar (o 3Themax cualquier 0. m ⎝uniformpérdidas 1000 de /s 4 ⎠The carga mwhich combinada /sen2del velocity.25 5-22 ⎛ kg/m 1Vuelva kJ/kg 3. elect. la the agua Defina 1 toturbine deducción kW the fluid Se at y presión b) Noting the atmosphere la free inlet. ventilador converted to ⎜ will para the If youelare renovar 2 ⎟ +a stude /m 3 . Equations =α 1m and 222 = Vrespectiva- ⎜volumétrico are ρV α gh2L to+ the open gze2a+ ⎟ e windrate is of water and potential at un point are grande + Turbine 1+de + 1Generator = + + 2 + h turbine. 76 se useful decrease a de atm ) ofhp travésand the mantiene pumping este dur.4 en condiciones kW for of el la it 1a extremo lbm/ft diferencia to sushaft de work dicha es ft /s)(32.de Analysis la W5-78 y ¿Cuáles de El (a) cm loss. aire de ofun cuarto pumpde baño cuyas dimensiones son in2 this 5-27C5–29C¿Cuáles Whatson las tresthehipótesis stagnation remaining pressure? one-third importantes Explain iscanwasted how itque sebeesta-because of the inefficiencies flow velocities on the pressure the 2change and across the themotor.63 ++/(0. Se The combined descartan encuales- 5-18Cba hacia Defina sumergida arriba. the efici 1 at drives theη una generator. Also.072°F mecánica está del 3 aire (50 m) dea it delivers diferentes. and and thus it can the from pump. it isinvestigue flows determined through 636 warekW el de efecto by the of EESFIGURE actual multiplying de pump (o cualquier laP5–23E velocidad power consists thecan of power otro del theviento beprograma change generated generation tanquede y del inbydiáme- thethis potential este flow ti-windby at both turbine the points m ⎜ at efficiency. pe 2W =pump. kinetic y kJ/kg que energy.81 to de to theagua m/s cube )(30 P4yof −laPthe 5 m) ⎝ 3 = wind velocity.   D el   f luido   ( ρ Ec. and de thuspresión the The de uno energy a energy otro lado at of points de la1 at water and 2ing point are constante una po) is 1Solution pe 1flow? ebombainvestigue =Explain.19 0.Determine m 2 (a) Vuelva /s mech. estámaximum 25determinem The arriba. de termopar (unite mech. 3 + . total neración del de estanque potencia delde ríocuya ladel por bomba ríosuperficie unidad completo como de libre masa enel ese efecto y ellugar. and theu su suma frictional pump =kg/m es pump. de V 2 which ⎞ The 5-20 combined = Se (1500 va its turbine-generator kg/s)(0. depósito pump. combined a) a razón la eficiencia to de is25pumped total L/s. converts and ¿En que Supongamqué 800 V está = / que 2(62. de el curso del flujo? Explíquelo. reservoir 7un cmlago en el aeremainslado una de velocidad jercicios  from electricity constant.3 un apunta entrada ft ) ⎜ vers versi una it4efi- = (potential 1 . Noting that z⎜⎝the iscontrol 2to =Vze4. points tion (V ¿En ≅ difference qué V ≅ m==4. m 2 echanical spuesta: Solving energy de for 15 ΔP ft3/s.116 idad mechanical energy ofiswater. multiplying outthe − eexit mech.5 andde remains por 1¿Cómo m.70 de 2)(mecánica ) carga ⎝ = =lbf 1 2. z3 = z4. la significado isbetween determined fluid = the descarga m(de fluid e mech.14 deAlso.116 kg/s = = ⎛ pump 2 ⎞2−3 m = 1 . the respecto wind What 5-­‐20. línea ⎟3=¿En quéloss. Use el Soft.2 m. at determineThe ρ agua. Use el Soft. 83 por Limited ciento.W definition.1) kN/m = 289. above mente. using it without permission. 2 lbm/s   ( ) ( 2 m/s ) ⎞ eglect losses in fluid era Δtes P using de chocar contra =(b) lospressures. de pur la analice misión ΔE ence su ymech.1 in 30 pe. = 0. gradiente kg/m ⎞ es 3 V y hidráulico? 30 30 melevación m ⎛ P es ¿En z.   bomba-motor a large de está W energía bombeando energy are yhite.727 inlet 750and or the 72. Substituting. descarga. .arede be hdetermined.072 =supplies kW of mechanical the Elwhich 800es ofucm to shaft Ework 5-74 la Analysis 2that 12masa ⎝ m/s. of a Discussionstream? a) energías. of this Elabore the wind mechanical unaturbine tubos tabla atde energy the los stated of resultados water cam- conditions. determine fromdeade70 lake la Analysis velocidad to en a nearby máxima Webychoose adepu 8 5-20 motor?lla que eficiencia combinada está en bomba-motor elbombear extremo agua de que dicha la manguera de la bomba o la deldirecto se apunta cenamiento 5-24 aSeoflabombea queseestá FLUID agua 20desde mMECHANICS un lago aagua razón hasta porun tanque L/s.toloss be =The E mech pump + E2mech loss.2 differ kinetic energy both points kW the water both deterlbm ylocations is essentially stationary. los 5-77A¿Qué boquilla de which factores hose rate. y la steadily 5-70Sigeneración -la gen salida Vuelva deade acertain potencia considerar potencia velocity. Bernoulli The es Fluye se is isthese de We 3overall connected la elevaagua 2 Vaceleración /2 two 25 per en points assume corrección the to unit inlet efficiency m un por the that en tubo andmass. to surfaces be de both point laof1energía 2.072 lugar.2% del aire flujo de flujo? la salida de laVcondiciones bomba = theW pump.07 ⎝ m) / 4 eficiencia 2 total del ⎠2 turbogenerador 4 A4 D 4 ⎝/ 10 2 4 tá ation 70ism750 kW. 2 = 1The . + bem W above determined.116 the 4(m−⎜⎜3m. en gz el 1curso (potencia and el efecto pe del en 2 la=Underground 0.7 kW:energy ofFIGURE water. ⎜ Solving LAS  3  Efor from NERGÍAS   ΔP andSON   normal substituting. 1 ) 370kN/m 2 = 289 . ⎠change ⎝ 1000 del kgyou across sistema If ⋅ m/s the de are ⎠ pump a 7. If you are a student using PROPRIETARY steady flow? MATERIAL. está 30demlosRespuesta: potencial factores arriba de ge.   Noting that z3 = z4. 2006 system. bombea the por elevation agua las desde álabes difference un sobre lago hasta la across un generación the pump de eólicaand alma. Substituting. exit2 of the pump. de gradiente 22 . Een la 305deducción free ismech m diámetro surfaces above = de of theρ del itW V la the tubo ecuación delivers gh storage underground dede topump. kWTome = of electric 0. mech W loss.uno 0 ⎞ possesses.81 m/s 2 )(30 ⎞m) m ⎝ 7. 4 incrementos se consumen de 5 20. flow por ciento 2 The gligible. lapérdida 2turbina 750 isatFIGURA energy its Desprecie kW a kinetic irreversible yare constant la 800 2⎝eficiencia 1000 kW negligible. would fluid give the =result ke2 = since flow energy the turbine -is equal to the at eleva-⎝ 32. pool at5–73 so- acon anada gi atTurbomáquinaria 6 η 2 la  diferencia Solution Water Ríois since pumped m/s from pressures at the inlet and outlet of the pump are W to be determined = comW ergy P/ρ both points bomba. elevation between u eleva p 20 causar que el agua de la manguera se eleve por arriba del nivel biosthe quiera enpérdidas la energía pool is por constant. pump are0). Vpo) 2 h re owerulic water generation turbine? po)e as Therefore. the f actual power Then en the incrementos can rate bepower. u = ηpool an. tankfree Also. kg/s 70 efficient kg/s)(0. which en is elevación⎝para = ⎜⎝6−.20 196 eólica m/s.81 presión = underground pump.25 electric kg/m power generation determined by multiplying the power pump. the =43The hose ⎝ to rise3 respectiva- 1theThe power windThe generador potential ne lablowing ischanges Wbomba. antes ofand y1the +pool. =pump. 500 m3/s Determine unen un alugar a) elvelocidad 90gasto m por máximo bios inenpiping. viento Discussion = gz W bomba.b)Elabore una tabla de los resultados y Infiere 20 m m ⎜ 3 + α 3 deEenergía? 3Thus. the (62. ¿Qué es E la es líneaP.4 of kW 20Aire amlarge de reservoir. 1 kJ/s Reconsider ⎛ Prob. We Wtake estática. presión atmosférica de 100 kPa.0. determined E mech.7 20. take the pump as the control volume. diámetro ⎝ 1000 m varía Se /s dedescartan ⎠20 hasta cuales- 80 m. 7. 12 del 4 = hp) The 1de programa the P fluido 25The2 L/s. take 1031DkW theviento | 2density 1031 kW ofsoplando water to uniformemente be por ρ = unidad 1000 kg/m 2 3. 1000shaft. este Solution !!"# Wind is blowing ⎝ 2 2steadily at a certain ⎠ !!"#$%&$'velocity. the temperatura temperature (o pump loss. álabes3 de ⎜ la(c) turbina. ismelout =potencial VA =be =ρde V = mkJ/kg (e=mech. 2tube la Solution Underground water giv determine the velocity which water be rbine-generator  5-19 arriba de efficiency Considere de agua and un the turbine efficiency hacia lago are to be una determined.  The major PMcGraw-Hill 3. the 2una reservoir potential 2rías eficiencia 3 es supplies of de total the 5 m.   conditions. /s hidráu- rates Noting of water that become z = z duce . eficiencia ofde inthe esta si W kinetic outsteadily de wind laplanta.pumped kPa. inlet en nozzleel and = W flujo outlet m is observed − estacionario? ⎜ of +E the α pump to + rise are gz 25 to ⎟ . de the se 5-30C these puede ⎜Defina two points medir.4687 elevation!!"#!"# difference across 5-76effects the pump Seare usaware negligible? isuna Properties debomba negligible. The mechanical energy of air per incompressible level (z 1 = 0). de fluid y it analícelos.velocities enters tanto at both the tur. and energía the del ofsuelo. con max álabes . de dicha bine nozzles1 atThe Assumptions 700flow kPaisabsolute steady and withincompressible. m) velocidad 2 ⎛a razón 1 kJ/kg del de viento ⎞ L/s. Determine 2 (a) mech. V = 2 = ⎞2 =inefficiencies 3 es de 5 m. uwater. and the ! lake remain constant.070enmla /sdirección proportional and⎜ 1 to heads. incom.470 Properties Frictional 2ExplainkPa We L what y2en take440 losses pump.Energía). superficie presiones libre y está c) cargas. +EW = m gz =2pump . volume noise unid Assumptions Then the rate rrección 1 The de la elevations energía cinética of the tank son despreciables.Wind The 5-23I overall Unefficiency sensor deof termopar turbine the combined diferencial pump-motor en la admisión unitaisrazón determined   líneas la Then the from mass coinciden and its1Substituting. tanque 64 pump . = The =−1031 kW ) =points ( pe1is− zero =través 5-76 SeEl 15usa flow una through bomba Pde a=mediante control 7P hp ). the ft arriba es frictional ρ del effects 2 Assumptions 1 e 1 =1keThe = de elevations potencia = eléctrica ⎜ tank es 2 and de 2 2 the = 0determi- ⎟ lake . de the por lake turbina unidadto4belado de point blecen 15-27C and the agua. in 2 mpower generation 1 = VΔP potential líneas by50 the coinciden m u ⎝ con la superficie efficiency.4 la lbm/ft 5-73 ) ssumptions arriba 1 The ba elevation de sumergida la superficie ofdethe del 3 kW reservoir lago. subsuelo be fondo de Discussion pipping kg/L tres =del 0. m) a lake ⎜= varía 29 la . (zsistema=determined 0).incompressible We takederrección point 1 de la 1 ofat the 2 energía free ρ hp both surface cinética1000 points of correction son underground ≅3 V2 ≅ water.116 =ftmhasta × 10 +(1000 una kg/m m /s + gz 1 . la which densidad energy to is del the V /2 per turbine. the energy equation fo2 kPa ≅ 289 kPa uni.Analysis The power Tanque the since corresponding Ean = Ein useful water energy P/ρ at potencia both points is pump =is P2negligible.1 kW f⎟ 1000 m /s kg⎠ 5–65C What pump.14 studenis ne ir per unit Therefore. los × 10 factores −3 13and both m de 2co- /s are open to ráulica? anical energy the free rate of pump of water . (0. en4 donde 86 + W turbine m/s se genera + E mech V 3.in hidroeléctrica. (que0(. to motionless.cinética.qué V di. mass derivation Ecuación ofen the deBernoulli energíaequation? achers and se educators una partícula thefor pump course el must flujo preparation. se Tome apunta :he wind Then turbine the 5-21 -gengenerador   | ΔE mech. © actual gz 5-71I. . conditions. = es de 5lugar shaft.68 a generar 7 energy potencia kJ/kg)and eléctrica the turbine a2 través efficiency de la instala. ba sumergida Exprese In 3 (a)Underground The The the de ladepump-motor effect absence 3 ecuación kW of yof dethe cona drawskinetic turbine. are open duce to theincompressible 5-68C 5–67C de atmosphere UndergroundWhatnivel hasta is del (Pwater the flow8= cm 1agua 750kinetic through en 2is kW un pumped atmtanque energy a Further. estacionario? negligible. = ( Respuesta: − 4 . the la the kinetic differenc pum η = η rías elect.86 m/shead . quiera above pérdidas Suponga at a rate porque of 70 L/s la velocidad while consuming varía de 20. relaciona 3⎜ kW)so. and hydrostatic pressure. determine el gasto volumétrico h L = h pump.same 452 (ke1iskg/s eléctrica 0) sisince se thesupone theat5-28C free atsurface W(b) Defina of ηinlet presión =the pool. este ti- energy theof free se puede surface medir.4 potencia kJ/s kW por ⎛ 1 kPa ⋅ m ⎞ bomba 3 es de 27 fera? gz 2 ⎛ P (9. 1 +fl m/s 1000 m 290/sm2 ⎠ pressible fluid. mVeffects the the pump-motor pump cinética tank. the take efficiency only the | silake form 1031 of the of surface kW mechanical combined as the reference pump-motorenergy thelevel wind unit 1ispossesses. la bomba.000 atthe a gal/min rate fanagua of is 25 ain V desde=L/s narrowA2Vby either 2 un= cross-section a( lago 10-kW D direction.6 m /s)(9. (a) In water from accordingly. unit is horizontal soflowthat3rate z = of constant 3along /s.centerThen the of energya sharp-edgedequation for10-cm-diameter this control volumeorifice. de 0. 0.90 15 .135 /s)(9. 4 The effect of the kinetic energy another as 2above the V = 0 and P = P = α P . where c disponible! 0.6 monly comprar /h to e tube. en la noche.0 ) fan section reduces to = 3.106 'final ! 75 rad/s higher 2 the Zto2 =heat 25 because ft turbine remaining gráficahalf is de converted los resultados toIfheat the y because energy of air while analícelos.7 oil by be kPa Wfan. u 2 u2 ⎝ 1 kg 2 N ⋅ m/s 1sión graduated ⎜flow aelloses (other ⋅ m/s a 20. Because suministrado manometer the tube por toisbe la bomba so1. the turbine head is determinedof imperfections.0336=m waterPft.2Section W (1) the flowing 8 m/s condition? ⎝ Pump– 4 ⎠ pump. flow fan–motor (V+ 12 ≅ ratesWe reemplazar ⎛Eis 0).5 P4 W − P3 = = by sióna mercury nar. Se usa una bomba subir agua hasta una altura de 15 m.4 W.035 © 2006 P3. laminar ! (32"LV)/(#gd2).2 ! m g entrada negligible. the electric power = P 0 P3. The temperature of the ingre of air thewhereremaining pipe half diameter is converted is 90 to mm.0375 kg/s)(1.49flowm/sloses in ) 2 kW − (12.152 bombear (a) 85. atwo la sides1of ⎛ consists enters boquilla Nathe ofestáafantank. gal).discharging pérdida © 2006 The Assumptions irreversible of 0. deliver mediante water the reducer. 3 mminimizar ! para to whose Properties La velocidad 2 40la2m vibración dimensions ⎠ ydel el The areaire2density ruido. hasta unaaBecause 25-cm- altura de moto The must pressure drop corresponding to a differential friction height head of has 1. privadas Fig. at dores a system que usenis h la ! cQ 2 capacidad . 2 ρg V 22 2 g + α5–83 = respectively. ciné- (V1 u= =0). Siexit la3 area and better recovery. pump pressure 3 and that the fluid diameter velocity at the free surface3of the tank is v Vthis mains efectos 1 same ⎟ nada de50laThe P unidad V ventilador-motor specified amount que semass usará puede tomarse V + gz1 +of comonoise.2 is converted tois the mechanical Pump. 1 kg ⋅ m/s ⎠⎝ 1000 N ⋅ m/s 2 ⎠ and educators course preparation. Determine pressurized to Utilities The 0. de Si z3 =ulaz4 pérdida 7 m.25loss.5 loss = E Amech Exhaust water Ventilador column loss.4 VW= A V 2 Pump =8 (m/s D 22 / 4 )V 2 D2 = =1 = 0.08 h2/m5. efficiency).2 m) W ⎜ ⎟ duran ⎜cross-section ⋅ m/s ⎟⎠⎝ diameter ⎠lion barrels per day (bbl/day) (1 exits bbl !through 42 U. boquilla Se que conecta está The en total una head lossdeisaldicha el manguera extremo 6. pump in this case and W pump. long turbine vertical capillary through la cual Generator aumenta tube.2 Wand surface 2 at reduces the2to is inlet unknown.88 head loss is tempera.tank Notingair pres- that the elevation el volumen de aire of2en 10 min.102 h2/h1 ! %&2& #&12&[1 1 teachers and educators for course Therefore. 50 = (0. D 1along8the m/s 1 8 m/sP3. If the total V 22 irreversible 1 head loss in 27 m ctor? Is it Wfan. steadysumidero and 3 es incompressible. output 4 The of entracorrection a effectEn of the Substituting kinetic energy and noting correction factors isthat Wturbine. what If the elevation (3) of 240 ft tothe3a turbine.137 0.fan 2 voirs to restore the situation. anWaldo  con. 3. is measured cuyo diámetro toWbe 470 kPa 5-443 )(0.156 (a) 102 kPa. la depósito unadetermine bomba que está the (potencia 25 m mechanical en(c)arriba.=78% ⎝ is ⎟ = 55 kW 3.81 m/s )(11. ElSenivel conecta del una en un tanque manguera PROPRIETARY fondo del mtanque arriba del la P3.98 flow 1 The due ft rate flow/s P2 =toenters the of is   steady fan-motor and incompressible.154 h ! 0. Respuesta: Solution Z1 = 150 78.05. the loses of theAire place fan unit. pressurea small determine dropdiameter. determine the maximum Discussionvolume It flow appearsrate that of this hydroelectric turbine will generate 55 kW of electric power Turbine given under conditions.2 wattage generate are sistema by a el also de ft/s pump. S Para aprovechar esta oportunidad. the upper Pis 4 required reservoir = an inair daytime ⎜ ventilating ⎟ = 40fan to produce = 40power to 2produce Pa for a P3. presión (b) de the uno 2 a diameter otro ladoof de the rise 90 ft from the ground. 0. 1 − 3 . bre la eficiencia demecánica de m.4 W.-diameterFIGURA hasta stream city.6 ⎟ than 3/s and he éporpuedela ground.123 each pump.140 permitted only to precio alto. where a cQ is 20 de mc ! carga above 0.Lizcano.134 los durante 4500 pe-hp pipeagua and discharges italthrough está a20 nozzle. sistema determined. you aredeusing it⎝ without permission. equation alongturbine-generator is open to thethe atmosphere.120 (a)más 22 ft/s.116 kW.560 the pressure over the kg/m .correction energy Turbine in m.81 m/s 2 ) permitted 5-81 s de co. que what una horsepower compañía generadora motor is required 3. 3. If the pump efficiency is 75 un flui.plants = ofhose that Determine A 83will a bathroom per- by replacing be el connected used to the ⎜ bottom of a rate of air tankonceQis ⎟ m every ⎜ equippedwill result? minutes with while ⎟ = pressure a nozzle 435at lbm/s the end pointing straight up. fanfriction. entre take lasat points parties 1superficies and thelibres Assumptions 2 on a high price.92 only dV/dtto ! gh/(L # de co. gasto Vuelva determine only a 3 de agua si la pérdidaa short velocity considerar the minimum time el during η remains problema irreversible flow peak below turbine-waterrate gen gha specified 5-71I. Para Compilado  total e jercicios   turbogenerador d e   E nergía:   C enguel.25 ⎛ 1N ⎞⎛ 1 kW ⎞ 2 PROPRIETARY MATERIAL . u P is =4 = 2.5 ft. heat the because pres- Substituting into Eq.re- 78(1000 kg/m Companies.4 of the fan (V is 1 selling lbm/ft isSolution piping negligible. what if the irre.135 82 por ciento. to an el. length to produce power for a L. We take The(1) point to a elevation large 1 at3 the free difference surface of water in the2 tank. laabove reducer. reductor. drawscompañías seawater Inc.110 Z 2for=(a)25 ft 0. eficiencia turbina.134 A2 36-in-diameter 8 m/s since E mech. 2 pressure drop. reser-kg/m above 3 the](9. respectively. P3. 120 ft/s Whatefectos 2 is con- m/s nada the ⎞ in para air de since minimizar lamech.136 deseando permitted a submerged 5. pipe. If the pump efficiency 1640 is 75 hp teachers and educators for course preparation.03gasto m100 /sdekW agua ing m ofto ρpurchase(1.08 h 2 de /m 5 3. 370generation lbm/s 100 kW the entire ⎛volume 25. e . de baño flow effects one ratecuyas are ventilador negligible reservoir Point ⎛ofPTo is2ft. of the kinetic the de los⎜ electric diameter energy pipefactores power correction de rating ⎟ ⎜corrección of the factors. ciento. and 3 +W η 3 fan. for P3. pumping which is stations accounted every for by 10themi along thethe efficiency). diameter is reduced from 15 cm the exit to 8velocity cm byisareduced reducer. Limited distribution3.0V =⎠ determine percent. The tank = (0. the 18 mcasing =considerar available == 5must ft W3 be Respuesta: power 5-78elect /sAturbine. .25de ⎞ corrección kg/m = Patm and ⎛the flow velocity 1 W 3 and disregard ⎞de la A energía is negligible hpump. ghturbine. you system are antesisahin student y irreversible f! después 2 using delthis Manual. energy Esti.10 9478 Btu/s ⎞ air erated. pump. a investigate and considerar 2 educators el theforeffect problema course 5-74. © The McGraw-HillP3. 5-73 [ Also Respuesta: solved using 370 EES lbm/s on enclosed DVD ] 72 percent Water 2 2efficient and is driven by a 500-W motor. uP= + z 2 + h turbine. en deldonde se genera   pump. de tuberías − P10 =es Noting kW. u Wfan. cm the hastamide8tank.123. α = 1. Supongadeque la Trace pérdida 3 3 2 m)⎜⎜ ⎟ hat only halfdeofcarga de5–76 carga varía the electricA 7-hp 0(shaft) energy hasta pump 15 consumed enused is incrementos by tothatraise the fan-motor water 1 m. +Point +1histurbine.134 12Athe.flow 037 ft /s 20 L/s⎞⎛30/h. of water= from = the = 1. Wfan.3 m. and the Determine exit of the the turbine. 5-73 theSe debe seleccionar 8 m/s uidoincom- an en between points $0.68 Respuesta: de m corrección flow rate Q de m /h 3 la will energíaresult? cinética como 1.137. minimum respectively. 3.Limited The distribution central Answer: The power 0. turbine steady -gen m and e = η turbine-permitted incompressible.83 la(1000 kg/m )(0.136 A pump is toque deliveren lawater at 20°C y. The total head loss is 6.138Students m) m/ =of ⎠ 4 6. bypressure large 3 and 40 For lake Pa rise abefore4totowill design be need adischarging reser- flow on rate to it.138 boqui. which must be delivered to the oil by ( D 22 / 4 )V 2 D 2 = The P3.4 13 Vacross W. where V lake is the level. increase =nozzle 18 m3.motor Nbe 1. Bernoulli. H⎛ P5-78 1 N ⎞⎛ 1 kW ⎞ sión. and water level 808 height Answers H to Select supo ps 100 2 4 V . a hydroelectric volume bathroom andatmass determine night is V flow elatpower = much rates gasto 2 m plant. Properties theP5. línea the a pond.=u W pump −ρVE=mech related to del ventilador the fan–motor y c) la unit diferencia to be de purchased. also theeither two sides direction.25D2m= /s D1y .08/kWh thelarge be determined. 9dV/dt = 11 . thewe pumped pressure efficiency restore   the takefrom larger ofofpoints the a fan. determine 10 × 2 a)s(2) la⎜⎝ 1negligible. reductor.term (b) 3. The electric power cons output − P3 = 5-75 Vuelva ⎞teachers software.25 de 240 kg/m 3 ftuna hasta una . evated tank.en Analysis incrementos Note onlyde half1ofm. thatfan. central razón Takeesand de dethe tank the demanda energía eléctrica L suele ser mucho ATERIAL. m m is to the 2× 3 m rise).V [ 2 #2 sure P is 120 kPa. P3. u =m 0. to a very low level.   White. tuberías We ft. determine 72 la pérdida percent irreversible efficient and is de carga driven by en a éste. by the fan-motorthe corresponding unit is converted power ⎝ output ⎠ mechanical 3. A Discussion pump moves Note thatwateronly horizontally half of the at electric a rate energy of 0. 2 for 6 m /s + gz 2 ⎟ + W turbine + E mech. Use W irreversible el Soft- fan. pump50= m Wshaft powerpor 2ciento.128 .a!un it?2006 la energía pipeproducidaand discharges durante el díaQ por it through empresas a nozzle.142 de servicio to drive (a) público 1150 it? gal/min vende energía eléctrica a 0.03/kWh the exit 1 at and free surfaces night andtounis ventilador 2 reduces of the willing source topara The net electric 1from renovar paypower and el is tofor output $0.2 pump-turbine theVm) minimum ⎜ the average iscasing system value. (b) 1 ducemine 0. of preparation. you are using it without permission +$gd4(H # as in3. sink ispara 36 1 Pump 2rease on the inlet5. 1N todo 1the Wpoints⎞ stream 2 o se re- el used volumencan bede ρ Wfan. equation a través for de section reduces to water máximo across the Upstream thewhile pump ft. motor 0.6 ft lo. la Toflecha) a pump efficiency determine razón de the de 10whereof pressure kW. de carga is turbine del value.94 without h ! permission.137.loshose factores is connectedde corrección to the bottom de la energía of the tank.136 A pump is to deliver water at 20°C from a pond to an el- entrauna con a 5-71I desde En una una planta alturaThe 10 ratesft de 240 generadora of ftwater hasta are una hidroeléctrica.122 V Prob.equation for del the fan pipe agua.   m above   the e =inmínimo −bathroom = is −V36 = tas a la during the day than it hturbine.1. 3.137 ft) ⎝ height 1ABtu/lbm of fireboat the water draws ⎠ ⎝ jet is 1 kW seawater measured. the de 25 L/s.de density ⎛ 2inside- los P3.123 5-73and Water Se sides aire is to debe of de un be Therefore.   20 m agua *P3.133 A nética? (b) For air meanWhat velocity to remain 50 ⎛ P ft below the V 1specified ⎞ value. ft/1000 of pipe.percent 5. 3. efficiency shown 83 in Chap. D 2 the pump are 3 negligible /s. P eléc- potencia ! $ Qr * [r 2* 5-77 If theFluye pressure agua atenthe un centerline tubo horizontal. ⎝ P3.124 and the 41tankr/min kPa y 440If kPa. the day from de private pipe and discharges it ∆P through = 2 kPaa nozzle.5 mente. (b) 55 apun 3. que la antes isCompanies.misión andis 5-74If the depósito Se irreversible está está bombeando head agua loss desde aofrazón unpiping theenergy lago grande system hasta is 7the un m. 5–78 determine The water la level pérdida tank. pump thisheightcase andm/s) of W pump. de tuberías 5-79 Respuesta:esUsingde m /778. ⎞ L =0 (to Determinepump. 49 system m/s is 0. u ⎝= mα 2 aire taken 2 en to 10 2 be min.000 to be máximo do thedeeleva- removed.83(32.146 hel!día. (1). inefficiency) 2 Friction FIGURE are for this Air P5–81 motor is required to drivefriction between the water and air as well as in the h 25 luidokg/men )(0 de. The pump is 1 m above the pond.5aire como1 and 2 on the ρde 1.z 2Also. 500-W Tomemotor. V gal). 1.2pump.2=W 1 + h pump. línea pump. irre- 4 since the fan la 148 kN/m 2 is a narrow ⎛ 1000 17kgft. required periods.25 +α a kg/m otro 11 3lado gzy1 des- 2+the de +fan.25difference =electricity.02 consumed m /s. stopwatch and graduated cylinder.⎠Inc.2 long.158 (a) 169. 3 erm can be 5-72E cent. −obtained upper 1000 )value. of4 the the The low electric density power of air). zthe (1) and h Ldel utility 240 companies 204 often máxima potencia de Analysis 83 por eléctrica. difference lapipe a través pérdida Answer: across de irre. pumpflowing Substituting.03 m /s 2 ⎜ 3 and⎟ choosethe la water 1 ⎟ = 1. máximo de agua.4 como Wde Puno 1 ρgacross 1.5 3 versible +W percent de (potencia carga = mdel 4sistemala flecha) horizontal de P line.81 m/s 2 ) ⎜ bine and (b) ⎟ delivered by 9the pump.in V required the × 3pipes level and 2Substituting. throughbuta control to be 13. by the Vfan-motor (8unit m/s)is converted to the mechanical t.03 mW McGraw-Hill it. u = mα determine   2 the 2 initial 1 discharge We velocity 1. Take head  the loss kinetic between 1 and 2. fan/motor ⎟ = 1 . nerar 100 kW de electricidad. the del1. canDetermine increase to as la much salidaasde 923. generator ble. 1 V A u 4 ( 0 .68 head respectiva. 3. 81 ⎟ = )( 71 . ρ de mecánica versible eficiencia del ρ cargaReconsider sistema de la bomba.03ρ m Therefore. The pump is 1 m above the pond. head and com- for course 470 preparation.1=Wηso- Plot flow the is ⎠ Inc.de ⎟ 1 ⎟2+ =baño WdelPatm selected pump Assumptions cuyas =.035Substituting. Deter. si Pump Vla=pérdida kg/m 3 )(0across irreversible Answer: power m the .4 kPa.124 A 34-hp motor through a ⎛long pipe is given hf. A we(8take 2gh )It D is 2 points m/s) planned⎜ (0. at a higher power ⎛ P1produced elevation as effect indicated of⎞inthe during kinetic Fig. it pumps water from lower to upper reser- edmisión dicha that discharge voirVSe 3 = z45. you are using it 3.069 known.03 m 3head /s ⎝ 1 W on the ware Water ⎠ at mechanical 20˚C EES (o cualquier otroofprograma P4 − Pprograma the1. aia eebom-surface a 24-in. the betwo supplied to of 4(0sides the the upper overcome 3 of experiment. 20 m 2(distribution Limited 3. un empresario considera 3.6 m /scie delfunction agua está ofabierta temperature. Noting that z3 = z4 and V3 2= V4 pump. ⎠dem agua se andeleva a stopwatch.2 W 1 N ⋅ m/s ⎛ the ⎞ rating   N/mof the V −2 P3P5-73 by (8 m/s) 70 na bom. WTome c) la=la(c) y elect fan. water ft volume between these two points t 10pipe the lower ⎟⎠ como ⎜ ⎞⎛de frictional por ciento. = thenegligible 5-84 Determine Assumptions flow (other than = 0).025) distribution están from3.0 at t ! 70 5-75 ⎜ loss Vuelva ⎟ =dea40 N/m = efficiency considerar   40 Pathis el problema Then the netiselectric turbine 5-74. with an 2.of unidad entiremotor-ventilador volume in que airReservoir 10 debe min. you h2respectiva-5 are pump usingisit without pañíaspermission. The 0. what system horsepower controlwithvolume a known flow D rate.0375 kg/s)(1. unfor cuyo at 20°C diámetro respectively. andthe flow z =(or.236-in-diameter pipeline on acarries oil (SG ! 0. gasto 4V P5–80 FIGURE 4(0. flow (or. u 27 diameter m the por tube arriba pipe. a =15-m to electric unit 0. the extracted turbine head and the mass and volume flow = 1.138 Companies. because of Determine air).5 ρm. Then = aconstant that z surface forthe hydraulic energy percent. the moved seleccionar the elevation 2fan. m and inlet 435 de1exit la lbm/s Thefuente flow ofisythe sides el fan. por ©P3 (or ciento (1000 2006 kg/m The other) McGraw-Hill P4 W1fan. It is planned to ity édel to puede se nivel remain to rise below the specified mate the value. T Δt 3 ft10 × 60 s 2 the fan-motor⎜unit. The of the fan water theonfriction density a at this3. of will.   free surface isM 20unson.0 = Lake ρ = kg/m )( 9 . the ⎛ P diameter   V of 2 ⎞drains 2the fan⎟casing should 3 to thebeatmosphere. +α2 = 7 m 0 .7 of kPa.81 represents m/swhereThe 2 of)(1. cinética and the como nozzle 1. should be P3.casing 7 kN/m to 2 should the = 68. potencia kgthe⋅el m/s ⎟⎝de la⋅ m/s ⎠ equation 1 Nenergy oamount se re. casing.0375 kg/s)(1.2. Further.25 kg/m is 3 .and diferencia densidad the pressure = delde presión aire=difference 2. Bernoulli.89) 7atm1corresponds mil.manguera If and the educators del pump tanque se efficiency yforlacourse apunta is 75 preparation. h loss.5ftmpor The ciento + available = head of a hydraulic − = u turbine = and u its overall efficiency are given. air situation.104 ⎠ ' ! (%Ve /R) ln gráfica varía deelevation. If generación the pump 1. m   If you are P ρ of this hydroelectric 4 a P student 3 using m / ρ turbine this Manual. the pipe.89) theftVfan a at30-cm- 1 mil- 2 m/s power rating of(c)the To fan/motor determine theunit friction Air pressuremust head difference be= is loss 2. trica de Inc. horsepower the diameter which W of the be fan delivered = 1. 6determine percent. suelo.138 . Atm night. fluye ergy 5-71Ifactors this una planta isturbine. 4 The 2of least far6. 5 Mass. ⎝on1000 a kg ⋅ m/s reservoir diameter in the ⎠ daytime d. En la shown línea que in Fig. and(b máximo whose de agua. 5–74. m making /s eficiencia el in gasto a mecánica volumétrico horizontal the exitvolumétrico pipe diameter of the pipe equalcon to the 3 de la bomba es deP3.betoThe be21000 tankkg/m mα 2 incompressible coverand is the flow density of mercury airtight. is at night. f e el gas. in el the the aumento flow turbine remains mínimo is measuredlami- de pre. the kW (work %15.in increments otro 3 = de⎜ este ti-⎟ = 40 N/m = 40 Pa raise the pressure head loss po)PROPRIETARY e investigue of air vary by from el 40 efecto MATERIAL Pa 0 to before 15 m de la. velocity E unidad not loss = E latovibración isDiscussion ventilador-motor exceed 8yNote pump ininvolved. energy The take correction loss 2of We take advantage factors available The flow the density this negligible. is steady αof= ft and water 1. 3.135 1 kN ⎞ inby 4 m.118 ⎟ = 163 P! kW $QuV   n(cot . as in Fig. The fan 0.08en the /m éste. Δdetermine effects Section 60 (a) are b) diámetro wattage 2 ⎠ the of Cengel 05. ge- 3 rs the tur. generadora el agua P3. unit Also.136 15kPa agua cm Ahasta before en pump unand istubo 8teachers cm to after horizontal.25 / 4(1000 13N toandm) 2to1be 3 ⎟⎛⎜ 5–84del / 4Wm/s tanque ⎞2Water P3.loss the = 8 . Az 2 large tank is initially hfilled e with laciona este último. Para (a)determineuna The volume eficiencia el of gasto air 1total turbogenerador necesario =para 2m × 3ftm × 3 m = 18 m . where electric powercapacity is gen.0 carte el efecto ) Therefore.2 (a) (other extracted than 27m/s) m 2 by the tur- entrada fan. sincemin the fan ispump.05. This demonstrates the need a larger turbine subir de 15 m. 1Vmust dimensiones water because to another renovar sufficiently be son of el ⎞ isat the low density3 of air).25 unit ismhorizontal its overall so that turbine–generator constant flow effi- (or. you are using it without permission. Discussion y mechanical analícelos.112 redesigning disponible T the ! ṁR turbine 2 0 ' and la agua de5–77 bomba hasta Water es de una flows82alturapor at aciento. mP3.25 8 kg/m . de delalapérdida Pump- bomba. The ⎝carga 3 turbine /s Companies. P b) el diámetro V P V Wufan. When the Trace the turbine la energy consumed head available. system in Fig.5fondo teachers ft. evated Siirreversible se cm que 5-44 The head mediante la pump loss in presión isun en 1them reductor.α Also. u = W pump factor 3 − Etomech a pressure at thethe loss. A hose is connected to the bottom of the tank at the ground correction across the turbine is negligible.kN ⎝ Companies.025) from ! minimum The a submerged pressure rise supplied by ne blades. ufor=steady equation up. ρg 2 2g sufficiently L far from the turbine. la pre-a edirecto 2de extracción terpor la stream Aire W m α Δ= P(0. at the theend required the flow ofConstant-hose rate of water is pointed 1 straight 2 the 8-in. =2 1P water e +h + 2 m above (1) and hpump. irreversible Suponga que de carga la pérdida so. andit?efficiency.130 500-W motor. 1 de1 m. . day. the elevation because ciency of effects the is 78 are low negligible density percent. +Vα22 Si Then Si el 2efficiency + gz ⎟ of theventilador volume electric ofand Properties the 2 ⎟ + W turbine low debe power. directo ahaciavery . Water it. and ⎠the (SG 1.2 W 68 .9°. que(bh . α elect = η turbine -gen mghturbine. losses power while D = 2 overcoming in for by turbine friction elocity un flui. which h is = accounted for by the efficiency). 25 the Si L/s. section Sicylinder. × 60 s tuberías entre las superficies libres de la fuente y el sumidero es Chapter 5 Mass.0375 carga is to del be determined. a la The atmósfera. inver Wmfan.this 5–82E loss control 2 = mα 2 level fan..035 dem315 /s. and the orifice tica. the A noche a freeprecios surface más is 20bajos m above con elthe finpump. preparation. Downstream bine Penand (b) of the delivered by the W the pump. 5-76water.05. The tank is at sea tencia y72evitar percent laefficient construcción and is de driven nuevas by a plantas 3. del The such suelo.135 1.is negligible !3m m 5–73 (Vaire ⎜ !m1⎜3=A de m.pipeline turbine carries oil V 2in(SG ! (8 0. − z1 P2 = the Patm.   system Chapter is hfBernoulli.96 d 2Z/dt2 # 2gZ/L presi 0.144 y26 kW desea 5-82 pagar 0. Noting that z = 20 to generate a27 m and z = an incom- antiene (2) is loss reservoir ! 61V # 2 ft # s . and point 2 at the top of the water trajectory wh un nd flui- mantiene 2 reduces mm/s ! 3⎝ m. determine e V la 3 )(9. and the tank on una nsitya of stas la air  desde is given unato altura be 1. lower ofmínimo × air3 through m water × prices 3the necesario m flows to 18 casing = Ing.133 factores fan to 1 . efficiency of imperfections. V =) If 18them Analysis includes a) 3 fan la ⎜ the potencia to pump replace We = 0.03370 produced 3 /s)de fan lbm/s pump are to bekg/s = during 0.135 draws theinterpreted pipe.S. device El tanque to measureestá a the nivel viscosity del marofywater as a la superfi- aiba he del anboqui- -ground. For a design flow rate of which energy que accounted offor by 17the 25 m arriba.10For miaalong 2agua. asMATERIAL in yFig. 2 The available only eto head remains constant. inética? 2 ⎜ 1 kgbetween ⋅ m/s 2 ⎟⎠⎝ 1 N ⋅Substituting.135 = pump-turbine FIGURE P5–73 Therefore.5 2 entrepreneur ft /s being is Analysis pumpedand Constant- ρ considering from thus h sections = 20. Point 12isFrictional sufficiently Thefar total losses from the head fanloss is 6.3those = the 6. en to el dondebe agua fluye Properties5–80 The density demand of air for is given electric to power be 1. negligible.the fan unit. generadoras de servicio d público venden la energía en la w rate of mente. fan losses cuarto 1 Also. atm/s) ifand= 0. ofunit theloss take air energy the through todo to be density the casing of water must Wfan.137. W de 5-44⎛ 1N ⋅ m/sthe este ti- ⎞ 2 1 ne la ware de EES Substituting (o cualquier . Wfan.2eficiencia W ⎛ 1 N ⋅ m/s mecánica de 1la bomba. f ! 0. efficiency energy associated ρ opportunity. Analysis el (a) m/s this que that ruido. average factors pumping is velocity negligible. a given In reality moto For air to remain below the specified diameter the fan ⎜ 2 ⎟ ae hasta pressure iameter dichahpof difference 5–74 across Therefore. ⎝ at 1000 rate kgchorro reduces of to thermometer.133 m pero 3.069 m draws 6. la eficiencia if a much total larger exitturbogene- de 3.el 03 turbine–generator m problema is toefficiency fanpower Solution ventilate 5-71I. )(0+. rise ofcondition? orifice to be 1. 5-72E For avoid Vuelva overall building a Solution new m 0expensive . generate incompressible. to case 2 hThe se and the La usará minimizepipe W volume effect eficiencia getof 1 puede of vibration u air= W 2 frictional combi- 1atm tomarse inpump the −and 2 E mech bathroom losses isfor =.03 dólar/kWh. to Use elLet power output Soft- V becomes Water levelZ1 = 150 ft 3.   from encourage must param 3 be ge- .68 m fireboat draws seawater (SG ! 1. u ρbe determined.2 m loss in the is 13 mercury ft/1000 ft of pipe.21 mso y truir un depósito grande 40 m arriba del nivel del lago. of the u fan. conduce barat directo 2 =unit. Si la eficiencia de 7 hp (potencia that almost en W lahalf flecha) of turbine mecánica= η the para available turbine m gh pressure turbine = η head turbine ρ is Vdiscarded gh turbine as kinetic energy. = and V determine 3 =uV fan. determined turbina. ⎞ una hydraulic abomba. and Energy Equatio and Energy 0.108 (a) V ! V0 /(1 # 5-76 Se usa una bomba de 7 hp (potencia en la flecha) para 82 percent.08 dólar/kWh por la potencia producida durante 3. and 5-74 25 m 3 = V4above está since bombeando 15. tica. Inc.4 place pumping stations everym. =than 6.148 hcons. 5-77andde duce Fluye 440P3.137 surface is open to the atmosphere. kg/m pump m = z0.qxd sure in terms (gage) 2/22/06of heads 2 that for will 6:29 thecause AM a Page turbine water 223 reduces stream to from the nozzle to unidad motor-ventilador que debe comprarse. t comprarse. 2 ! cQ . Problems 203 Equations and Energy un flui. de alentarThe head a3. m aire ⎜ Then to ⎜ no 2dimensiones +α debethe ventilate 1 a+building energy 2 sobrepasar son bathroom⎟ gz 2 ⎟ +equation2 W8turbine + Efor mech. lion barrels per day (bbl/day) fan(1 bbl ! Ventilador V 42 U. and should bespeed ⎟ through Fig.122 a pond to L lasanthe ! %h1 rate. 3esta turbina. © 2006 turbineThe =se0McGraw-Hill .simple arriba. m⎝noLa of3air !debe L is msobrepasar eficiencia diameter ! 3 given to⎠beThe combi- 3water m. boquilla que estápercent.1352 negligible. + 3 P gz isvelocidad in to be pipes.07 de ism /s po- 3 los factores0.at 3. Generator Note 2 3.9and cm volume flow rate Q is measured with the bomb igible. /s)fan-motor −ρthe P2 pump = (= ρ Hg must − ρcm supply gh/ aρminimum m V pressure mate the It rise horsepower willofbe68. 6 fte + h L 1m (1. ρ   3 The loss 62. el-flow (cot () 470 m3The kinetic kPa /s. 3electric system is horizontal Water Theso Ppower between water atm. 1000 All desired kg/m V 2 inelectric 1 2. confrom 3. 3 ■deFigure 36 5–72E 3 Reconsider P5.9 in the fluid mechanics laboratory at Penn State use riba del = = 2.30 3 2 *P3. along is = 5–71E. If the pump is c !Equations eras an.114 su diameter (a) 414 is used r/min.Pelect =manometer fan. the 5-43 ⎜ energy equation ⎟ for+ (1 the . a narrow2 and neglecting fan. that the laminar the net head loss electric la pé FIGURA 1. Substituting . negligible. volume of de 83 airThen inpor sell the the Inpower ciento. 3   o de de la tube. The head loss in the potencia eléctrica.03 air it m  mean having 3 pumps Discussion /s ⎝avelocity W ⎠from 1water uniform speed =The lower of[(13 40 result . Tome la ηdensidad We take points del 0. combined ⎜ required.125 zde.5 so that P 1 ft.126 losloss in consumi. viscometer.0375 Suppose the irreversible kg/s the flow avelocity utility headcompany negligible.)(204 and the P3.98from /sthe totofan be ρ so=that theP 3 dur.81 Limited m/s 2 )(85 m) distribution ⎜ ⎜ permitted only ⎟⎜ ⎟ to 3. IfTome ground.0375 kg/s P − P2 α (V12 2− V 22 ) elaciona del esteventilador fan último. Esti- el nivel ■ Figure P5.137 The McGraw-Hill riodos A pico. u the loses of 4the fan3unit. = 2 D32neglecting kN/m un head =loss 148=kPa is flow losesare 0(other . determine el gasto uninlet gasto diameter. from a razón a sepond deter-re. turbine the fan Properties will If you raise the arepressure aWe studentof take airusing by this the40density Manual. es de moutput loss in the permitted of a hydraulic alta only turbine durante to elisdía proportional tonoche the available turbine frecuencia. 1 1 2 2 +α2 3 3 2 /s) + .560 to ft/s. inside-diameter ⎛ 2piping mech loss. we take points water 3 andis4gal/min on in to bemeasured. 5-79 the powerUna turbina output can hidráulica be increased tieneto 85 74 m kWde by carga 3.distribution Limited gen ρVgh turbine. © 2006P3. hidroeléctrica. 5-78 suelo. 5–75 PROPRIETARY Prob.025) from a submerged do el degas-la 5-78 El nivel A del agua en un tanque está 20 m arriba del costosas flow que ratesólo 5-49 Q m3utilizarán se /h will result? un tiempo corto 3.fireboat Estas Companies.113 ft ) lbf. . Ifde you are a student   5-80 using this LaManual.069 cmWhat water from is thethe friction tank. rate of determine 0.and to ras an- mass f 100flow nerarof100 kPa. a la líneaentrance losses areturbine– decombined negligi.yenergy Si McGraw-Hill 440 sefreemide surface correction kPa. (SG también Limited ! 1. 03 m /s) =city. rates airsumers kW through V =an deV to use electricidad. where Chapter   5 Mass. the end of the hose is pointed straight up. un αLa +0).138. Estimate the the power in kW (a) extracted by the tur. )V 2 (shaft) 2 / 4grande the =friction 148 and hasta pump. the remainingfrom determined ofhalf pump is is converted of imperfections. turbine. the points where 1 and 2 reduces m/s2 ⎠ 40 m to water P3. each pump. The horizontal line. 1tokgthe⋅ m/s 2 ⎟⎜ 1000 N ⋅ m/s Eff. de 0 los resultados hasta 15 m.α we take the reference level at the bottom of the tank. fan so that D = 6 in ρ water 1 g 2g ated to the carte Take el efecto the (8 air m/s) 8-in.   5–71E máxima ecity.03 αVSubstituting. volume WThe u water 1 inthe turbine a tank is are 66 accounted ft above the ground. For example.100 (a) 507 m/s and útil d po) elainvestigue breresults. = 0.S. Determine Water thethe the fan aerodynamic voirs fantoraise is will being unit. u is the friction ⎜ m⎜ + α 1 ρ 1head loss 2 between hV ⎟ + gz1 ⎟ + W pump = m⎜ = V 1 = = and 27 − V⎜ 2.MATERIAL EES . FIGURA 3 .2 unit energía W must be ciné- 2. 03⎝2 m /s) D1 / 4 (0. Suponga percent. Pa before of discharging water you arebeusing to ρ 20˚C = 1000it without kg/m permission. the V1cuarto un respectively. el ventilador (8 debe m/s) reemplazar mech. presión 20 yfactors después m to above be Inc. = 2 ⎛= =on the nozzle de extracción ⎞4P3. en el extremo what horsepower de dicha motor PROPRIETARYmanguera is required seMATERIAL apunta to drive. W diameter ! " # " gene oet andla the una de m bomba 78. If you are a student using this Manual. and eficiencia elmecánica discuss efecto them.25 inlet g la kg/m 1 and exit 3 2 gthey effect sides des.   If you are a5-53 student using this Manual. generadoras 34 what kW Respuesta:P3. the diameter of⎟ the fan casing. se genera usually much higher evated tank. the theminimum amount valueloss. cross-section⋅ m/s Estimate 2 and⎞ neglecting the(8power. Discussion en lareductor. 15. radorthe espower de 78generation por ciento. At (b) P5-73 of night. 20ρ 2 in m.9 water ) from water fan/motor unit must be 2. Bernoulli.144 TheChapter 5 Mass.shaft = η motor W electric = (0. Noting that z = z .4 − 17.08 m) 2 / 4 Motor V V 0. how much horsepower does the turbine develop? P3. which is 0.141.73. to travel a maximum horizontal distance. System friction Solution age velocityAinpump is pumping the pipe.u = ρVghpump. 5-55 .144 hpump. If head losses are negli. P3. what Theismechanical horsepower efficiency needed to drive it? of the pump is to be determined.05. where V is the average velocity in the supply V V 0.05[ (8.144 and Energy creates a 20°C water jet Equations oriented 5-86 are approximated by hf ! 27V2/(2g). resulting in a pump per.4 m)⎜⎜ 2 ⎟⎜ ⎟⎝ 1 kN ⋅ m/s ⎟⎠ = 25.  Munson.813 = 0. theThe pump is measured.81 m/s 2 ) ⎜⎝ 1N ⎟ ⎠ 2(9. 3 All the losses in the pump are accounted for by the pump efficiency and thus hL = 0.5 kW Wpump. head rise lossesofare oil6.140 Steam enters a horizontal turbine at 350 lbf/in2 absolute.5 lbm/s. respectively. P3. shaft 31. The mass flow is 2. where V is the aver. as in Fig. u = ⎜ ⎟+ = 47.145 The large turbine in Fig.0 = 30. u = (860 kg/m 3 )(0.1 m 3 /s V2 = = = = 8. 3 Properties The density of oil isz1 given D = 10 cm 15 m Analysis = 6 in2 at the inlet and the exit of the pump. 4 The kinetic energy correction factors are Jet given to be α1 = α2 = α = 1. oil atis a75specified If the pump rate.6 kW ⎝ 1000 kg ⋅ m/s ⎠ Then the shaft pumping power and the mechanical efficiency of the pump become W pump. P3.Pump u ρg 2 2g   L ρg 2g  P3. T1 = 20°C Tank : = 200 L   580°C. De = 5 cm 25 m = 50 ft to be ρ = 860 kg/m . P3.6 kW ηpump = = = 0. P3.9 m/s A1 D1 / 4 (0.142.       saturated conditions.5V /(2g).000 N/m 2 ⎛ 1 kg ⋅ m/s 2 ⎞ 1.143 P3. the energy We take points 1D and 1 2 equation for the pump reduces to 2m Pump P1 V12 P2 V 22 P2 − P1 α (V 22 − V12 ) P3.145 diverts the river flow un-   EESwheretion rate.813 = 81.141 + α + z + h = + α + z 2 + hturbine.Ing. The pressure percent effi. jet may and the be approximated by motor the efficiency is specified. varies with the flow rate. and the P1 = 1500 kPa   heat losses are 7 Btu/lb of steam.84 m/s ) 2 − (19.81 m/s 2 ) ⎛ 1 kN ⎞⎛ 1 kW ⎞ W pump. u = + ρg 1 2g 1 pump.3%     Wpump.84 m/s A2 D 22 / 4 (0. What power must be de- livered by the pump? Assumptions 1 The flow is steady z 2 = 150 ft and incompressible.1 m 3 /s)(9.81 m/s 2 )(30.5 kW   Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies. System friction18losseskW are hf ! formance curve as in Fig.142 A typical pump has a head which. trajectory of frictionless particles.   W aldo   L izcano. EES der a dam as shown. pump in Fig.139 Air supply: Valve P3.  White. Suppose 3 that this pump 2 2 3.12 m) 2 / 4 1 Substituting. Pump D1 = 9 cm D2 = 3 cm P3.9 m/s) 2 ] hpump.1 m /s Pump V1 = = 2 = = 19. cient.141 Water at 20°C is pumped at 1500 gal/min from the lower to the upper reservoir. 2 The elevation difference across the pump is negligible.90)(35 kW) = 31.9 × 0. the useful pump head and the corresponding useful pumping power are determined to be Oil 400.4 m (860 kg/m 3 )(9. Pipe friction losses P3. for a given shaft rota.5 in m. u 25. e + hP3.     gible. and 12 ft/s and is discharged at 110 ft/s and 25°C Compilado  ejercicios  de  Energía:  Cenguel.